Collaborative Statistics (Custom Online Version Modified by T. Short)

1. Additional Resources

2. Author Acknowledgements
3. Student Welcome Letter

4. Sampling and Data

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. Introduction

. Statistics

. Probability

. Key Terms

. Data

. Sampling

. Variation

. Answers and Rounding Off
. Frequency

. Summary

11.
12.

Practice: Sampling and Data
Homework

5. Descriptive Statistics

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. Introduction

. Displaying Data

. Graphing Qualitative Variables
. Histograms

. Stem and Leaf Displays

. Box Plots

. Measures of the Location of the Data

. Measures of Central Tendency

. Skewness and the Mean, Median, and Mode
. Measures of the Spread of the Data

12,
13,
14.

Summary of Formulas
Practice 1: Center of the Data
Practice 2: Spread of the Data

15.

Homework

6. Linear Regression and Correlation

. Introduction to Bivariate Data
. Linear Regression and Correlation: Linear Equations
. Linear Regression and Correlation: Slope and Y-Intercept

of a Linear Equation

. The Regression Equation

. Correlation Coefficient and Coefficient of Determination
. Guessing Correlations Simulation

. Prediction

. Linear Regression and Correlation: Summary

. Practice: Linear Regression

. Homework

7. Probability Topics

1.
. Terminology

. Independent and Mutually Exclusive Events
. Two Basic Rules of Probability

. Contingency Tables

. Venn Diagrams

. Tree Diagrams

. Summary of Formulas

. Practice 1: Contingency Tables

. Practice 2: Calculating Probabilities

. Homework

12.

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Introduction

Review

8. Discrete Random Variables

1.
2:

2:
4.

Introduction

Probability Distribution Function (PDF) for a Discrete
Random Variable

Mean or Expected Value and Standard Deviation
Common Discrete Probability Distribution Functions

CONI OD UI

. The Binomial Random Variable
. Binomial Experiments
. Summary of the Binomial Formulas
. Practice 1: Discrete Distribution

9.
10.
11.

Practice 2: Binomial Distribution
Homework
Review

9. Continuous Random Variables

. Introduction

. Continuous Probability Functions

. Introduction to Normal Distributions

. The Standard Normal Distribution

. Z-Scores

. Normal Distribution: Areas to the Left and Right of x
. Areas of Normal Distributions

. Calculations of Probabilities

. Summary of Formulas

. Practice: The Normal Distribution

. Homework

12.

Review

10. The Central Limit Theorem

. Introduction

. Introduction to Sampling Distributions
. The Central Limit Theorem for Sample Means (Averages)
. The Central Limit Theorem for Sums
. Using the Central Limit Theorem

. Summary of Formulas

. Practice: The Central Limit Theorem
. Homework

. Review

11. Confidence Intervals

1.

Introduction

Deviation Unknown, Student's-t

. Confidence Interval for a Population Proportion
. Summary of Formulas
. Practice 1: Confidence Intervals for Means, Known

Population Standard Deviation

. Practice 2: Confidence Intervals for Means, Unknown

Population Standard Deviation

. Practice 3: Confidence Intervals for Proportions
9.
10.

Homework
Review

12. Hypothesis Testing: Single Mean and Single Proportion

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15.
16.

. Introduction
. Null and Alternate Hypotheses
. Outcomes and the Type I and Type II Errors

. Distribution Needed for Hypothesis Testing
. Assumption
. Rare Events

. Decision and Conclusion

. Additional Information

. Summary of the Hypothesis Test
. Examples

12.
. Practice 1: Single Mean, Known Population Standard

Summary of Formulas

Deviation

. Practice 2: Single Mean, Unknown Population Standard

Deviation
Practice 3: Single Proportion
Homework

17. Review
13. Hypothesis Testing: Two Means, Paired Data, Two Proportions
1. Introduction
2. Matched or Paired Samples

. Homework
7. Review
14. The Chi-Square Distribution
1. Introduction
. Notation
. Facts About the Chi-Square Distribution
. Goodness-of-Fit Test
. Test of Independence
. Summary of Formulas
. Practice 1: Goodness-of-Fit Test
. Practice 2: Contingency Tables
. Homework
10. Review
15. Appendix
1. Solution Sheets
1. Solution Sheet: Hypothesis Testing for Single Mean
and Single Proportion

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Paired Data, and Two Proportions
3. Solution Sheet: The Chi-Square Distribution
2. Symbols and their Meanings
3. Formulas
4. Notes for the TI-83, 83+, 84 Calculator
16. Tables

Additional Resources

This module catalogs several of the resources available for teachers and
students using the Collaborative Statistics (col10522) textbook and its
derivatives. This module provides links to the complementary teacher's
guide, supplemental materials including video lectures and additional
problem sets, accessibility information, collection version history and
errata, and a list of related works and teachers who have adopted them for
their courses.

Additional Resources Currently Available

e Glossary

e View or Download This Textbook Online
¢ Collaborative Statistics Teacher's Guide

e Supplemental Materials

e Video Lectures

e Version History.

e Textbook Adoption and Usage

e Additional Technologies and Notes

e Accessibility and Section 508 Compliance

The following section describes some additional resources for learners and
educators. These modules and collections are all available on the
Connexions website (http://cnx.org/) and can be viewed online,
downloaded, printed, or ordered as appropriate.

Glossary

This module contains the entire glossary for the Collaborative Statistics
textbook collection (col10522) since its initial release on 15 July 2008. The
glossary is located at http://cnx.org/content/m16129/latest/.

Below are links to additional resources:
Link to the Statistics Glossary by Dr. Philip Stark, UC Berkeley

http:// statistics. berkeley.edu/~stark/SticiGui/Text/gloss.htm

Link to Wikipedia

http:// http://www.wikipedia.org/
(Search on "Glossary of probability and statistics.")

View or Download This Textbook Online

The complete contents of this book are available at no cost on the
Connexions website at http://cnx.org/content/coll0522/latest/, Anybody can
view this content free of charge either as an online e-book or a
downloadable PDF file. A low-cost printed version of this textbook is also
available here.

Collaborative Statistics Teacher's Guide

A complementary Teacher's Guide for Collaborative statistics is available
through Connexions at http://cnx.org/content/col10547/latest/. The
Teacher's Guide includes suggestions for presenting concepts found
throughout the book as well as recommended homework assignments. A
low-cost printed version of this textbook is also available here.

Supplemental Materials

This companion to Collaborative Statistics provides a number of additional
resources for use by students and instructors based on the award winning
Elementary Statistics Sofia online course, also by textbook authors Barbara
Illowsky and Susan Dean. This content is designed to complement the
textbook by providing video tutorials, course management materials, and
sample problem sets. The Supplemental Materials collection can be found
at http://cnx.org/content/col10586/latest/.

Video Lectures

e Video Lecture 1: Sampling and Data

e Video Lecture 2: Descriptive Statistics

e Video Lecture 3: Probability Topics

e Video Lecture 4: Discrete Distributions

e Video Lecture 5: Continuous Random Variables

e Video Lecture 6: The Normal Distribution

e Video Lecture 7: The Central Limit Theorem

e Video Lecture 8: Confidence Intervals

e Video Lecture 9: Hypothesis Testing with a Single Mean
e Video Lecture 10: Hypothesis Testing with Two Means
e Video Lecture 11: The Chi-Square Distribution

e Video Lecture 12: Linear Regression and Correlation

Version History

This module contains a listing of changes, updates, and corrections made to
the Collaborative Statistics textbook collection (col10522) since its initial
release on 15 July 2008. The Version History is located at
http://cnx.org/content/m17360/latest/.

Textbook Adoption and Usage

This module is designed to track the various derivations of the
Collaborative Statistics textbook and its various companion resources, as
well as keep track of educators who have adopted various versions for their
courses. New adopters are encouraged to provide their contact information
and describe how they will use this book for their courses. The goal is to
provide a list that will allow educators using this book to collaborate, share
ideas, and make suggestions for future development of this text. The
Adoption and Usage module is located at
http://cnx.org/content/m18261/latest/.

Additional Technologies

In order to provide the most flexible learning resources possible, we invite
collaboration from all instructors wishing to create customized versions of
this content for use with other technologies. For instance, you may be
interested in creating a set of instructions similar to this collection's
calculator notes. If you would like to contribute to this collection, please
use the contact the authors with any ideas or materials you have created.

Accessibility and Section 508 Compliance

e For information on general Connexions accessibility features, please
visit http://cnx.org/content/m17212/latest/.

e For information on accessibility features specific to the Collaborative
Statistics textbook, please visit http://cnx.org/content/m17211/latest/.

Author Acknowledgements
This module contains the author acknowledgements for the Collaborative
Statistics textbook/collection.

For this second edition, we appreciate the tremendous feedback from De
Anza College colleagues and students, as well as from the dozens of faculty
around the world who taught out of the first and preliminary editions. We
have updated Collaborative Statistics with contributions from many faculty
and students. We especially thank Roberta Bloom, who wrote new problems
and additional text.

So many students and colleagues have contributed to the text, both the hard
copy and open version. We thank the following people for their
contributions to the first and/or second editions.

At De Anza College:

Dr. Inna Grushko (deceased), who wrote the glossary; Diane Mathios, who
checked every homework problem in the first edition; Kathy Plum, Lenore
Desilets, Charles Klein, Janice Hector, Frank Snow, Dr. Lisa Markus, Dr.
Vladimir Logvinenko (deceased), Mo Geraghty, Rupinder Sekhon, Javier
Rueda, Carol Olmstead; Also, Dr. Jim Lucas and Valerie Hauber of De
Anza’s Office of Institutional Research, Mary Jo Kane of Health Services;
and the thousands of students who have used this text. Many of the students
gave us permission to include their outstanding word problems as
homework.

Additional thanks:

Dr. Larry Green of Lake Tahoe Community College, Terrie Teegarden of
San Diego Mesa College, Ann Flanigan of Kapiolani Community College,
Birgit Aquilonius of West Valley College.

The conversion from a for-profit hard copy text to a free open textbook is
the result of many individuals and organizations. We particularly thank Dr.
Martha Kanter, Hal Plotkin, Dr. Judy Baker, Dr. Robert Maxfield of
Maxfield Foundation, Hewlett Foundation, and Connexions.

Finally, we owe much to Frank, Jeffrey, and Jessica Dean and to Dan,
Rachel, Matthew, and Rebecca Illowsky, who encouraged us to continue

with our work and who had to hear more than their share of “I’m sorry, I
can’t” and “Just a minute, I’m working.”

Student Welcome Letter
Dear Student:

Have you heard others say, “You’re taking statistics? That’s the hardest
course I ever took!” They say that, because they probably spent the entire
course confused and struggling. They were probably lectured to and never
had the chance to experience the subject. You will not have that problem.
Let’s find out why.

There is a Chinese Proverb that describes our feelings about the field of
Statistics:

I HEAR, AND I FORGET
I SEE, AND I REMEMBER
I DO, AND I UNDERSTAND

Statistics is a “do” field. In order to learn it, you must “do” it. We have
structured this book so that you will have hands-on experiences. They will
enable you to truly understand the concepts instead of merely going through
the requirements for the course.

What makes this book different from other texts? First, we have eliminated
the drudgery of tedious calculations. You might be using computers or
graphing calculators so that you do not need to struggle with algebraic
manipulations. Second, this course is taught as a collaborative activity. With
others in your class, you will work toward the common goal of learning this
material.

Here are some hints for success in your class:

e Work hard and work every night.

e Form a study group and learn together.

e Don’t get discouraged - you can do it!

e As you solve problems, ask yourself, “Does this answer make sense?”
e Many statistics words have the same meaning as in everyday English.

¢ Go to your teacher for help as soon as you need it.

e Don’t get behind.

e Read the newspaper and ask yourself, “Does this article make sense?”
e Draw pictures - they truly help!

Good luck and don’t give up!

Sincerely,
Susan Dean and Barbara Illowsky

De Anza College
21250 Stevens Creek Blvd.
Cupertino, California 95014

Introduction
This module provides a brief introduction to the field of statistics, including
examples of how these topics shows up in a variety of real-life examples.

Student Learning Outcomes
By the end of this chapter, the student should be able to:

e Recognize and differentiate between key terms.
e Apply various types of sampling methods to data collection.
¢ Create and interpret frequency tables.

Introduction

You are probably asking yourself the question, "When and where will I use
Statistics?". If you read any newspaper or watch television, or use the
Internet, you will see statistical information. There are statistics about
crime, sports, education, politics, and real estate. Typically, when you read a
newspaper article or watch a news program on television, you are given
sample information. With this information, you may make a decision about
the correctness of a statement, claim, or "fact." Statistical methods can help
you make the "best educated guess."

Since you will undoubtedly be given statistical information at some point in
your life, you need to know some techniques to analyze the information
thoughtfully. Think about buying a house or managing a budget. Think
about your chosen profession. The fields of economics, business,
psychology, education, biology, law, computer science, police science, and
early childhood development require at least one course in statistics.

Included in this chapter are the basic ideas and words of probability and
Statistics. You will soon understand that statistics and probability work
together. You will also learn how data are gathered and what "good" data
are.

Statistics

This module introduces the concept of statistics, specifically the ability to
use Statistics to describe data (descriptive statistics) as well as draw
conclusions (inferential statistics). An optional classroom exercise is
included.

The science of statistics deals with the collection, analysis, interpretation,
and presentation of data. We see and use data in our everyday lives.

Optional Collaborative Classroom Exercise

In your classroom, try this exercise. Have class members write down the
average time (in hours, to the nearest half-hour) they sleep per night. Your
instructor will record the data. Then create a simple graph (called a dot
plot) of the data. A dot plot consists of a number line and dots (or points)
positioned above the number line. For example, consider the following
data:

po.70066.56,50.50.5775009

The dot plot for this data would be as follows:
Frequency of Average Time (in Hours) Spent Sleeping per Night

o

o Oo

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o 0 0 0 °o o o
5 6 7 8 9

Does your dot plot look the same as or different from the example? Why? If
you did the same example in an English class with the same number of
students, do you think the results would be the same? Why or why not?

Where do your data appear to cluster? How could you interpret the
clustering?

The questions above ask you to analyze and interpret your data. With this
example, you have begun your study of statistics.

In this course, you will learn how to organize and summarize data.
Organizing and summarizing data is called descriptive statistics. Two ways
to summarize data are by graphing and by numbers (for example, finding an
average). After you have studied probability and probability distributions,
you will use formal methods for drawing conclusions from "good" data.
The formal methods are called inferential statistics. Statistical inference
uses probability to determine how confident we can be that the conclusions
are correct.

Effective interpretation of data (inference) is based on good procedures for
producing data and thoughtful examination of the data. You will encounter
what will seem to be too many mathematical formulas for interpreting data.
The goal of statistics is not to perform numerous calculations using the
formulas, but to gain an understanding of your data. The calculations can be
done using a calculator or a computer. The understanding must come from
you. If you can thoroughly grasp the basics of statistics, you can be more
confident in the decisions you make in life.

Levels of Measurement and Statistical Operations

The way a set of data is measured is called its level of measurement.
Correct statistical procedures depend on a researcher being familiar with
levels of measurement. Not every statistical operation can be used with
every set of data. Data can be classified into four levels of measurement.
They are (from lowest to highest level):

e Nominal scale level
e Ordinal scale level
e Interval scale level
e Ratio scale level

Data that is measured using a nominal scale is qualitative. Categories,
colors, names, labels and favorite foods along with yes or no responses are
examples of nominal level data. Nominal scale data are not ordered. For
example, trying to classify people according to their favorite food does not
make any sense. Putting pizza first and sushi second is not meaningful.

Smartphone companies are another example of nominal scale data. Some
examples are Sony, Motorola, Nokia, Samsung and Apple. This is just a list
and there is no agreed upon order. Some people may favor Apple but that is
a matter of opinion. Nominal scale data cannot be used in calculations.

Data that is measured using an ordinal scale is similar to nominal scale
data but there is a big difference. The ordinal scale data can be ordered. An
example of ordinal scale data is a list of the top five national parks in the
United States. The top five national parks in the United States can be
ranked from one to five but we cannot measure differences between the
data.

Another example using the ordinal scale is a cruise survey where the
responses to questions about the cruise are “excellent,” “good,”
“satisfactory” and “unsatisfactory.” These responses are ordered from the
most desired response by the cruise lines to the least desired. But the
differences between two pieces of data cannot be measured. Like the
nominal scale data, ordinal scale data cannot be used in calculations.

Data that is measured using the interval scale is similar to ordinal level
data because it has a definite ordering but there is a difference between data.
The differences between interval scale data can be measured though the
data does not have a starting point.

Temperature scales like Celsius (C) and Fahrenheit (F) are measured by
using the interval scale. In both temperature measurements, 40 degrees is
equal to 100 degrees minus 60 degrees. Differences make sense. But 0
degrees does not because, in both scales, 0 is not the absolute lowest
temperature. Temperatures like -10° F and -15° C exist and are colder than
0.

Interval level data can be used in calculations but one type of comparison
cannot be done. Eighty degrees C is not 4 times as hot as 20° C (nor is 80°
F 4 times as hot as 20° F). There is no meaning to the ratio of 80 to 20 (or 4
to 1).

Data that is measured using the ratio scale takes care of the ratio problem
and gives you the most information. Ratio scale data is like interval scale
data but, in addition, it has a O point and ratios can be calculated. For
example, four multiple choice statistics final exam scores are 80, 68, 20 and
92 (out of a possible 100 points). The exams were machine-graded.

The data can be put in order from lowest to highest: 20, 68, 80, 92.

The differences between the data have meaning. The score 92 is more than
the score 68 by 24 points.

Ratios can be calculated. The smallest score for ratio data is 0. So 80 is 4
times 20. The score of 80 is 4 times better than the score of 20.

Exercises

What type of measure scale is being used? Nominal, Ordinal, Interval or
Ratio.

1. High school men soccer players classified by their athletic ability:
Superior, Average, Above average.
. Baking temperatures for various main dishes: 350, 400, 325, 250, 300
. The colors of crayons in a 24-crayon box.
. Social security numbers.
. Incomes measured in dollars
. A satisfaction survey of a social website by number: 1 = very satisfied,
2 = somewhat satisfied, 3 = not satisfied.
7. Political outlook: extreme left, left-of-center, right-of-center, extreme
right.
8. Time of day on an analog watch.
9. The distance in miles to the closest grocery store.
10. The dates 1066, 1492, 1644, 1947, 1944.

AuBRWN

11. The heights of 21 — 65 year-old women.
12. Common letter grades A, B, C, D, F.

Answers 1. ordinal, 2. interval, 3. nominal, 4. nominal, 5. ratio, 6. ordinal,
7. nominal, 8. interval, 9. ratio, 10. interval, 11. ratio, 12. ordinal

Glossary

Data
A set of observations (a set of possible outcomes). Most data can be
put into two groups: qualitative (hair color, ethnic groups and other
attributes of the population) and quantitative (distance traveled to
college, number of children in a family, etc.). Quantitative data can be
separated into two subgroups: discrete and continuous. Data is
discrete if it is the result of counting (the number of students of a given
ethnic group in a class, the number of books on a shelf, etc.). Data is
continuous if it is the result of measuring (distance traveled, weight of
luggage, etc.)

Statistic
A numerical characteristic of the sample. A statistic estimates the
corresponding population parameter. For example, the average number
of full-time students in a 7:30 a.m. class for this term (statistic) is an
estimate for the average number of full-time students in any class this
term (parameter).

Probability
This module introduces the concept of probability as a mathematical
measure of randomness, including a number of real-world applications.

Probability is a mathematical tool used to study randomness. It deals with
the chance (the likelihood) of an event occurring. For example, if you toss a
fair coin 4 times, the outcomes may not be 2 heads and 2 tails. However, if
you toss the same coin 4,000 times, the outcomes will be close to half heads
and half tails. The expected theoretical probability of heads in any one toss
is + or 0.5. Even though the outcomes of a few repetitions are uncertain,
there is a regular pattern of outcomes when there are many repetitions.
After reading about the English statistician Karl Pearson who tossed a coin
24,000 times with a result of 12,012 heads, one of the authors tossed a coin
996

2,000 times. The results were 996 heads. The fraction >555 is equal to 0.498

which is very close to 0.5, the expected probability.

The theory of probability began with the study of games of chance such as
poker. Predictions take the form of probabilities. To predict the likelihood
of an earthquake, of rain, or whether you will get an A in this course, we
use probabilities. Doctors use probability to determine the chance of a
vaccination causing the disease the vaccination is supposed to prevent. A
stockbroker uses probability to determine the rate of return on a client's
investments. You might use probability to decide to buy a lottery ticket or
not. In your study of statistics, you will use the power of mathematics
through probability calculations to analyze and interpret your data.

Glossary

Probability
A number between 0 and 1, inclusive, that gives the likelihood that a
specific event will occur. The foundation of statistics is given by the
following 3 axioms (by A. N. Kolmogorov, 1930’s): Let S denote the
sample space and A and B are two events in S . Then:

© 0< P(A) <1;.
e If A and B are any two mutually exclusive events, then
P(A or B) = P(A) + P(B).

Key Terms
This module introduces a number of key terms related to statistical
sampling and data.

In statistics, we generally want to study a population. You can think of a
population as an entire collection of persons, things, or objects under study.
To study the larger population, we select a sample. The idea of sampling is
to select a portion (or subset) of the larger population and study that portion
(the sample) to gain information about the population. Data are the result of
sampling from a population.

Because it takes a lot of time and money to examine an entire population,
sampling is a very practical technique. If you wished to compute the overall
grade point average at your school, it would make sense to select a sample
of students who attend the school. The data collected from the sample
would be the students' grade point averages. In presidential elections,
opinion poll samples of 1,000 to 2,000 people are taken. The opinion poll is
supposed to represent the views of the people in the entire country.
Manufacturers of canned carbonated drinks take samples to determine if a
16 ounce can contains 16 ounces of carbonated drink.

From the sample data, we can calculate a statistic. A statistic is a number
that is a property of the sample. For example, if we consider one math class
to be a sample of the population of all math classes, then the average
number of points earned by students in that one math class at the end of the
term is an example of a statistic. The statistic is an estimate of a population
parameter. A parameter is a number that is a property of the population.
Since we considered all math classes to be the population, then the average
number of points earned per student over all the math classes is an example
of a parameter.

One of the main concerns in the field of statistics is how accurately a
Statistic estimates a parameter. The accuracy really depends on how well the
sample represents the population. The sample must contain the
characteristics of the population in order to be a representative sample. We
are interested in both the sample statistic and the population parameter in
inferential statistics. In a later chapter, we will use the sample statistic to
test the validity of the established population parameter.

A variable, notated by capital letters like X and Y, is a characteristic of
interest for each person or thing in a population. Variables may be
numerical or categorical. Numerical variables take on values with equal
units such as weight in pounds and time in hours. Categorical variables
place the person or thing into a category. If we let X equal the number of
points earned by one math student at the end of a term, then X isa
numerical variable. If we let Y be a person's party affiliation, then examples
of Y include Republican, Democrat, and Independent. Y is a categorical
variable. We could do some math with values of X (calculate the average
number of points earned, for example), but it makes no sense to do math
with values of Y (calculating an average party affiliation makes no sense).

Data are the actual values of the variable. They may be numbers or they
may be words. Datum is a single value.

Two words that come up often in statistics are mean and proportion. If you
were to take three exams in your math classes and obtained scores of 86,
75, and 92, you calculate your mean score by adding the three exam scores
and dividing by three (your mean score would be 84.3 to one decimal
place). If, in your math class, there are 40 students and 22 are men and 18

are women, then the proportion of men students is a. and the proportion of

women students is 3 Mean and proportion are discussed in more detail in

later chapters.

Note:

Mean and Average

The words "mean" and "average" are often used interchangeably. The
substitution of one word for the other is common practice. The technical
term is "arithmetic mean" and "average" is technically a center location.
However, in practice among non-statisticians, "average" is commonly
accepted for "arithmetic mean."

Example:

Exercise:

Problem:

Define the key terms from the following study: We want to know the
average (mean) amount of money first year college students spend at
ABC College on school supplies that do not include books. We
randomly survey 100 first year students at the college. Three of those
students spent $150, $200, and $225, respectively.

Solution:

The population is all first year students attending ABC College this
term.

The sample could be all students enrolled in one section of a
beginning statistics course at ABC College (although this sample may
not represent the entire population).

The parameter is the average (mean) amount of money spent
(excluding books) by first year college students at ABC College this
term.

The statistic is the average (mean) amount of money spent (excluding
books) by first year college students in the sample.

The variable could be the amount of money spent (excluding books)
by one first year student. Let X = the amount of money spent
(excluding books) by one first year student attending ABC College.

The data are the dollar amounts spent by the first year students.
Examples of the data are $150, $200, and $225.

Optional Collaborative Classroom Exercise

Do the following exercise collaboratively with up to four people per group.
Find a population, a sample, the parameter, the statistic, a variable, and data
for the following study: You want to determine the average (mean) number
of glasses of milk college students drink per day. Suppose yesterday, in your
English class, you asked five students how many glasses of milk they drank
the day before. The answers were 1, 0, 1, 3, and 4 glasses of milk.

Glossary

Average
A number that describes the central tendency of the data. There are a
number of specialized averages, including the arithmetic mean,
weighted mean, median, mode, and geometric mean.

Data
A set of observations (a set of possible outcomes). Most data can be
put into two groups: qualitative (hair color, ethnic groups and other
attributes of the population) and quantitative (distance traveled to
college, number of children in a family, etc.). Quantitative data can be
separated into two subgroups: discrete and continuous. Data is
discrete if it is the result of counting (the number of students of a given
ethnic group in a class, the number of books on a shelf, etc.). Data is
continuous if it is the result of measuring (distance traveled, weight of
luggage, etc.)

Proportion

e Asa number: A proportion is the number of successes divided by
the total number in the sample.

e Asa probability distribution: Given a binomial random variable
(RV), X ~B(n, p), consider the ratio of the number X of
successes in n Bernouli trials to the number n of trials. P/= ~.
This new RV is called a proportion, and if the number of trials, n,
is large enough, P’ ~N(p, ®*).

Data

This module introduces the concepts of qualitative data, quantitative
continuous data, and quantitative discrete data as used in statistics. Sample
problems are included.

Data may come from a population or from a sample. Small letters like x or
y generally are used to represent data values. Most data can be put into the
following categories:

e¢ Qualitative
¢ Quantitative

Qualitative data are the result of categorizing or describing attributes of a
population. Hair color, blood type, ethnic group, the car a person drives, and
the street a person lives on are examples of qualitative data. Qualitative data
are generally described by words or letters. For instance, hair color might
be black, dark brown, light brown, blonde, gray, or red. Blood type might
be AB+, O-, or B+. Researchers often prefer to use quantitative data over
qualitative data because it lends itself more easily to mathematical analysis.
For example, it does not make sense to find an average hair color or blood

type.

Quantitative data are always numbers. Quantitative data are the result of
counting or measuring attributes of a population. Amount of money, pulse
rate, weight, number of people living in your town, and the number of
students who take statistics are examples of quantitative data. Quantitative
data may be either discrete or continuous.

All data that are the result of counting are called quantitative discrete
data. These data take on only certain numerical values. If you count the
number of phone calls you receive for each day of the week, you might get
0, 1, 2, 3, etc.

All data that are the result of measuring are quantitative continuous data
assuming that we can measure accurately. Measuring angles in radians

amr Tm 3m
6 PI 3 ’ BA ’ T 2: 4 2
carry backpacks with books in them to school, the numbers of books in the

might result in the numbers etc. If you and your friends

backpacks are discrete data and the weights of the backpacks are continuous
data.

Note:In this course, the data used is mainly quantitative. It is easy to
calculate statistics (like the mean or proportion) from numbers. In the
chapter Descriptive Statistics, you will be introduced to stem plots,
histograms and box plots all of which display quantitative data. Qualitative
data is discussed at the end of this section through graphs.

Example:

Data Sample of Quantitative Discrete Data

The data are the number of books students carry in their backpacks. You
sample five students. Two students carry 3 books, one student carries 4
books, one student carries 2 books, and one student carries 1 book. The
numbers of books (3, 4, 2, and 1) are the quantitative discrete data.

Example:

Data Sample of Quantitative Continuous Data

The data are the weights of the backpacks with the books in it. You sample
the same five students. The weights (in pounds) of their backpacks are 6.2,
7, 6.8, 9.1, 4.3. Notice that backpacks carrying three books can have
different weights. Weights are quantitative continuous data because
weights are measured.

Example:

Data Sample of Qualitative Data

The data are the colors of backpacks. Again, you sample the same five
students. One student has a red backpack, two students have black
backpacks, one student has a green backpack, and one student has a gray
backpack. The colors red, black, black, green, and gray are qualitative data.

Note: You may collect data as numbers and report it categorically. For
example, the quiz scores for each student are recorded throughout the term.
At the end of the term, the quiz scores are reported as A, B, C, D, or F.

Example:
Exercise:

Problem:

Work collaboratively to determine the correct data type (quantitative
or qualitative). Indicate whether quantitative data are continuous or
discrete. Hint: Data that are discrete often start with the words "the
number of."

I

SCOMNDAUARWNP

. The number of pairs of shoes you own.

. The type of car you drive.

. Where you go on vacation.

. The distance it is from your home to the nearest grocery store.

The number of classes you take per school year.

. The tuition for your classes

The type of calculator you use.

. Movie ratings.

. Political party preferences.

. Weight of sumo wrestlers.

. Amount of money won playing poker.

. Number of correct answers on a quiz.

. Peoples’ attitudes toward the government.

. IQ scores. (This may cause some discussion. )

Solution:

Items 1, 5, 11, and 12 are quantitative discrete; items 4, 6, 10, and 14
are quantitative continuous; and items 2, 3, 7, 8, 9, and 13 are
qualitative.

Qualitative Data Discussion

Below are tables of part-time vs full-time students at De Anza College in
Cupertino, CA and Foothill College in Los Altos, CA for the Spring 2010
quarter. The tables display counts (frequencies) and percentages or
proportions (relative frequencies). The percent columns make comparing
the same categories in the colleges easier. Displaying percentages along
with the numbers is often helpful, but it is particularly important when
comparing sets of data that do not have the same totals, such as the total
enrollments for both colleges in this example. Notice how much larger the
percentage for part-time students at Foothill College is compared to De
Anza College.

Number Percent
Full-time 9,200 40.9%
Part-time 13,296 59.1%
Total 22,496 100%
De Anza College
Number Percent
Full-time 4,059 28.6%

Part-time 10,124 71.4%

Total 14,183 100%
Foothill College

Tables are a good way of organizing and displaying data. But graphs can be
even more helpful in understanding the data. There are no strict rules
concerning what graphs to use. Below are pie charts and bar graphs, two
graphs that are used to display qualitative data.

In a pie chart, categories of data are represented by wedges in the circle
and are proportional in size to the percent of individuals in each category.

In a bar graph, the length of the bar for each category is proportional to the
number or percent of individuals in each category. Bars may be vertical or
horizontal.

A Pareto chart consists of bars that are sorted into order by category size
(largest to smallest).

Look at the graphs and determine which graph (pie or bar) you think
displays the comparisons better. This is a matter of preference.

It is a good idea to look at a variety of graphs to see which is the most
helpful in displaying the data. We might make different choices of what we
think is the "best" graph depending on the data and the context. Our choice
also depends on what we are using the data for.

De Anza College

@ Full Time o Part Time

Student Status
Pat Tire Pat Time

Pert Tire

De Anza Foothill

Foothill College

@ Full Time & Part Time

Full Time
28.6%

Part Time
71.4%

Percentages That Add to More (or Less) Than 100%

Sometimes percentages add up to be more than 100% (or less than 100%).
In the graph, the percentages add to more than 100% because students can
be in more than one category. A bar graph is appropriate to compare the
relative size of the categories. A pie chart cannot be used. It also could not
be used if the percentages added to less than 100%.

Characteristic/Category Percent

Full-time Students 40.9%
Students who intend to transfer to a 4-year educational 48.6%
institution

Students under age 25 61.0%
TOTAL 150.5%

De Anza College Spring 2010

Omitting Categories/Missing Data

The table displays Ethnicity of Students but is missing the
"Other/Unknown" category. This category contains people who did not feel
they fit into any of the ethnicity categories or declined to respond. Notice
that the frequencies do not add up to the total number of students. Create a
bar graph and not a pie chart.

Frequency Percent

Asian 8,794 36.1%

Black 1,412 5.8%

Filipino 1,298 5.3%

Hispanic 4,180 17.1%

Native American 146 0.6%

Pacific Islander 236 1.0%

White 5,978 24.5%

TOTAL 22,044 out of 24,382 90.4% out of 100%

Missing Data: Ethnicity of Students De Anza College Fall Term 2007
(Census Day)

E thnicity of Students

40.0%

35.0%
20.0%
25.0%
20.0%
15.0%
10.0%

5.0%

°* Asian Black Filipino Hispanic Native Pacific White
Amer. Islander
36.1% 58% 53% 17.1% 06% 1.0%

Bar graph Without Other/Unknown Category

The following graph is the same as the previous graph but the
"Other/Unknown" percent (9.6%) has been added back in. The
"Other/Unknown" category is large compared to some of the other
categories (Native American, 0.6%, Pacific Islander 1.0% particularly).
This is important to know when we think about what the data are telling us.

This particular bar graph can be hard to understand visually. The graph
below it is a Pareto chart. The Pareto chart has the bars sorted from largest
to smallest and is easier to read and interpret.

Ethnicity of Students

Bar Graph With Other/Unknown Category

Pareto Chart With Bars Sorted By Size

Pie Charts: No Missing Data

The following pie charts have the "Other/Unknown" category added back in
(since the percentages must add to 100%). The chart on the right is
organized having the wedges by size and makes for a more visually
informative graph than the unsorted, alphabetical graph on the left.

Ethnicity of Students Ethnicity of Students

Pacific Islander

Glossary

Continuous Random Variable
A random variable (RV) whose outcomes are measured.

Example:
The height of trees in the forest is a continuous RV.

Data
A set of observations (a set of possible outcomes). Most data can be
put into two groups: qualitative (hair color, ethnic groups and other
attributes of the population) and quantitative (distance traveled to
college, number of children in a family, etc.). Quantitative data can be
separated into two subgroups: discrete and continuous. Data is
discrete if it is the result of counting (the number of students of a given
ethnic group in a class, the number of books on a shelf, etc.). Data is
continuous if it is the result of measuring (distance traveled, weight of
luggage, etc.)

Discrete Random Variable
A random variable (RV) whose outcomes are counted.

Qualitative Data
See Data.

Quantitative Data
See Data.

Sampling

This module introduces the concept of statistical sampling. Students are
taught the difference between a simple random sample, stratified sample,
cluster sample, systematic sample, and convenience sample. Example
problems are provided, including an optional classroom activity.

Gathering information about an entire population often costs too much or is
virtually impossible. Instead, we use a sample of the population. A sample
should have the same characteristics as the population it is
representing. Most statisticians use various methods of random sampling
in an attempt to achieve this goal. This section will describe a few of the
most common methods.

There are several different methods of random sampling. In each form of
random sampling, each member of a population initially has an equal
chance of being selected for the sample. Each method has pros and cons.
The easiest method to describe is called a simple random sample. Any
group of n individuals is equally likely to be chosen by any other group of n
individuals if the simple random sampling technique is used. In other
words, each sample of the same size has an equal chance of being selected.
For example, suppose Lisa wants to form a four-person study group (herself
and three other people) from her pre-calculus class, which has 31 members
not including Lisa. To choose a simple random sample of size 3 from the
other members of her class, Lisa could put all 31 names in a hat, shake the
hat, close her eyes, and pick out 3 names. A more technological way is for
Lisa to first list the last names of the members of her class together with a
two-digit number as shown below.

ID Name

00 Anselmo

ID

01

02

03

04

05

06

07

08

09

10

11

12

13

14

15

16

17

Name
Bautista
Bayani
Cheng
Cuarismo
Cuningham
Fontecha
Hong
Hoobler
Jiao

Khan
King
Legeny
Lundquist
Macierz
Motogawa
Okimoto

Patel

ID

18

19

20

21

22

23

24

25

26

27.

28

29

30

Class Roster

Name
Price
Quizon
Reyes
Roquero
Roth
Rowell
Salangsang
Slade
Stracher
Tallai
Tran
Wai

Wood

Lisa can either use a table of random numbers (found in many statistics
books as well as mathematical handbooks) or a calculator or computer to
generate random numbers. For this example, suppose Lisa chooses to
generate random numbers from a calculator. The numbers generated are:

94360 .99832 .14669 .51470 .40581 .73381 .04399

Lisa reads two-digit groups until she has chosen three class members (that
is, she reads .94360 as the groups 94, 43, 36, 60). Each random number
may only contribute one class member. If she needed to, Lisa could have
generated more random numbers.

The random numbers .94360 and .99832 do not contain appropriate two
digit numbers. However the third random number, .14669, contains 14 (the
fourth random number also contains 14), the fifth random number contains
05, and the seventh random number contains 04. The two-digit number 14
corresponds to Macierz, 05 corresponds to Cunningham, and 04
corresponds to Cuarismo. Besides herself, Lisa's group will consist of
Marcierz, and Cunningham, and Cuarismo.

Besides simple random sampling, there are other forms of sampling that
involve a chance process for getting the sample. Other well-known
random sampling methods are the stratified sample, the cluster sample,
and the systematic sample.

To choose a stratified sample, divide the population into groups called
strata and then take a proportionate number from each stratum. For
example, you could stratify (group) your college population by department
and then choose a proportionate simple random sample from each stratum
(each department) to get a stratified random sample. To choose a simple
random sample from each department, number each member of the first
department, number each member of the second department and do the
same for the remaining departments. Then use simple random sampling to
choose proportionate numbers from the first department and do the same for
each of the remaining departments. Those numbers picked from the first
department, picked from the second department and so on represent the
members who make up the stratified sample.

To choose a cluster sample, divide the population into clusters (groups)
and then randomly select some of the clusters. All the members from these
clusters are in the cluster sample. For example, if you randomly sample four
departments from your college population, the four departments make up
the cluster sample. For example, divide your college faculty by department.
The departments are the clusters. Number each department and then choose

four different numbers using simple random sampling. All members of the
four departments with those numbers are the cluster sample.

To choose a systematic sample, randomly select a starting point and take
every nth piece of data from a listing of the population. For example,
suppose you have to do a phone survey. Your phone book contains 20,000
residence listings. You must choose 400 names for the sample. Number the
population 1 - 20,000 and then use a simple random sample to pick a
number that represents the first name of the sample. Then choose every
50th name thereafter until you have a total of 400 names (you might have to
go back to the of your phone list). Systematic sampling is frequently chosen
because it is a simple method.

A type of sampling that is nonrandom is convenience sampling.
Convenience sampling involves using results that are readily available. For
example, a computer software store conducts a marketing study by
interviewing potential customers who happen to be in the store browsing
through the available software. The results of convenience sampling may be
very good in some cases and highly biased (favors certain outcomes) in
others.

Sampling data should be done very carefully. Collecting data carelessly can
have devastating results. Surveys mailed to households and then returned
may be very biased (for example, they may favor a certain group). It is
better for the person conducting the survey to select the sample
respondents.

True random sampling is done with replacement. That is, once a member
is picked that member goes back into the population and thus may be
chosen more than once. However for practical reasons, in most populations,
simple random sampling is done without replacement. Surveys are
typically done without replacement. That is, a member of the population
may be chosen only once. Most samples are taken from large populations
and the sample tends to be small in comparison to the population. Since this
is the case, sampling without replacement is approximately the same as
sampling with replacement because the chance of picking the same
individual more than once using with replacement is very low.

For example, in a college population of 10,000 people, suppose you want to
randomly pick a sample of 1000 for a survey. For any particular sample
of 1000, if you are sampling with replacement,

e the chance of picking the first person is 1000 out of 10,000 (0.1000);

e the chance of picking a different second person for this sample is 999
out of 10,000 (0.0999);

e the chance of picking the same person again is 1 out of 10,000 (very
low).

If you are sampling without replacement,

e the chance of picking the first person for any particular sample is 1000
out of 10,000 (0.1000);

e the chance of picking a different second person is 999 out of 9,999
(0.0999);

e you do not replace the first person before picking the next person.

Compare the fractions 999/10,000 and 999/9,999. For accuracy, carry the
decimal answers to 4 place decimals. To 4 decimal places, these numbers
are equivalent (0.0999).

Sampling without replacement instead of sampling with replacement only
becomes a mathematics issue when the population is small which is not that
common. For example, if the population is 25 people, the sample is 10 and
you are sampling with replacement for any particular sample,

e the chance of picking the first person is 10 out of 25 and a different
second person is 9 out of 25 (you replace the first person).

If you sample without replacement,

e the chance of picking the first person is 10 out of 25 and then the
second person (which is different) is 9 out of 24 (you do not replace
the first person).

Compare the fractions 9/25 and 9/24. To 4 decimal places, 9/25 = 0.3600
and 9/24 = 0.3750. To 4 decimal places, these numbers are not equivalent.

When you analyze data, it is important to be aware of sampling errors and
nonsampling errors. The actual process of sampling causes sampling errors.
For example, the sample may not be large enough. Factors not related to the
sampling process cause nonsampling errors. A defective counting device
can cause a nonsampling error.

In reality, a sample will never be exactly representative of the population so
there will always be some sampling error. As a rule, the larger the sample,
the smaller the sampling error.

In statistics, a sampling bias is created when a sample is collected from a
population and some members of the population are not as likely to be
chosen as others (remember, each member of the population should have an
equally likely chance of being chosen). When a sampling bias happens,
there can be incorrect conclusions drawn about the population that is being
studied.

Example:
Exercise:

Problem:

Determine the type of sampling used (simple random, stratified,
systematic, cluster, or convenience).

1. A soccer coach selects 6 players from a group of boys aged 8 to
10, 7 players from a group of boys aged 11 to 12, and 3 players
from a group of boys aged 13 to 14 to form a recreational soccer
team.

2. A pollster interviews all human resource personnel in five
different high tech companies.

3. A high school educational researcher interviews 50 high school
female teachers and 50 high school male teachers.

4. A medical researcher interviews every third cancer patient from
a list of cancer patients at a local hospital.

5. A high school counselor uses a computer to generate 50 random
numbers and then picks students whose names correspond to the
numbers.

6. A student interviews classmates in his algebra class to determine
how many pairs of jeans a student owns, on the average.

Solution:

. stratified

. cluster

. stratified

. systematic

. simple random
. convenience

AuBWN PRP

If we were to examine two samples representing the same population, even
if we used random sampling methods for the samples, they would not be
exactly the same. Just as there is variation in data, there is variation in
samples. As you become accustomed to sampling, the variability will seem
natural.

Example:

Suppose ABC College has 10,000 part-time students (the population). We
are interested in the average amount of money a part-time student spends
on books in the fall term. Asking all 10,000 students is an almost
impossible task.

Suppose we take two different samples.

First, we use convenience sampling and survey 10 students from a first
term organic chemistry class. Many of these students are taking first term
calculus in addition to the organic chemistry class . The amount of money
they spend is as follows:

$128 $87 $173 $116 $130 $204 $147 $189 $93 $153

The second sample is taken by using a list from the P.E. department of
senior citizens who take P.E. classes and taking every 5th senior citizen on
the list, for a total of 10 senior citizens. They spend:

$50 $40 $36 $15 $50 $100 $40 $53 $22 $22

Exercise:

Problem:

Do you think that either of these samples is representative of (or is
characteristic of) the entire 10,000 part-time student population?

Solution:

No. The first sample probably consists of science-oriented students.
Besides the chemistry course, some of them are taking first-term
calculus. Books for these classes tend to be expensive. Most of these
students are, more than likely, paying more than the average part-time
student for their books. The second sample is a group of senior
citizens who are, more than likely, taking courses for health and
interest. The amount of money they spend on books is probably much
less than the average part-time student. Both samples are biased. Also,
in both cases, not all students have a chance to be in either sample.

Exercise:
Problem:

Since these samples are not representative of the entire population, is
it wise to use the results to describe the entire population?

Solution:

No. For these samples, each member of the population did not have an
equally likely chance of being chosen.

Now, suppose we take a third sample. We choose ten different part-time
students from the disciplines of chemistry, math, English, psychology,
sociology, history, nursing, physical education, art, and early childhood
development. (We assume that these are the only disciplines in which part-
time students at ABC College are enrolled and that an equal number of

part-time students are enrolled in each of the disciplines.) Each student is
chosen using simple random sampling. Using a calculator, random
numbers are generated and a student from a particular discipline is selected
if he/she has a corresponding number. The students spend:

$180 $50 $150 $85 $260 $75 $180 $200 $200 $150

Exercise:

Problem: Is the sample biased?
Solution:

The sample is unbiased, but a larger sample would be recommended
to increase the likelihood that the sample will be close to
representative of the population. However, for a biased sampling
technique, even a large sample runs the risk of not being
representative of the population.

Students often ask if it is "good enough" to take a sample, instead of
surveying the entire population. If the survey is done well, the answer is
yes.

Optional Collaborative Classroom Exercise

Exercise:

Problem:

As a class, determine whether or not the following samples are
representative. If they are not, discuss the reasons.

1. To find the average GPA of all students in a university, use all
honor students at the university as the sample.

2. To find out the most popular cereal among young people under
the age of 10, stand outside a large supermarket for three hours
and speak to every 20th child under age 10 who enters the
supermarket.

3. To find the average annual income of all adults in the United
States, sample U.S. congressmen. Create a cluster sample by
considering each state as a stratum (group). By using simple
random sampling, select states to be part of the cluster. Then
survey every U.S. congressman in the cluster.

4. To determine the proportion of people taking public transportation
to work, survey 20 people in New York City. Conduct the survey
by sitting in Central Park on a bench and interviewing every
person who sits next to you.

5. To determine the average cost of a two day stay in a hospital in
Massachusetts, survey 100 hospitals across the state using simple
random sampling.

Variation

This module discusses statistical variability within data and samples.
Students will be given the opportunity to see this variability in action
through participation in an optional classroom exercise. This module also
has a section that discusses Critical Evaluation.

Variation in Data

Variation is present in any set of data. For example, 16-ounce cans of
beverage may contain more or less than 16 ounces of liquid. In one study,
eight 16 ounce cans were measured and produced the following amount (in
ounces) of beverage:

15.6.16.1 15.2 14.8 15.8 15.9 16.0 15.5

Measurements of the amount of beverage in a 16-ounce can may vary
because different people make the measurements or because the exact
amount, 16 ounces of liquid, was not put into the cans. Manufacturers
regularly run tests to determine if the amount of beverage in a 16-ounce can
falls within the desired range.

Be aware that as you take data, your data may vary somewhat from the data
someone else is taking for the same purpose. This is completely natural.
However, if two or more of you are taking the same data and get very
different results, it is time for you and the others to reevaluate your data-
taking methods and your accuracy.

Variation in Samples

It was mentioned previously that two or more samples from the same
population, taken randomly, and having close to the same characteristics of
the population are different from each other. Suppose Doreen and Jung both
decide to study the average amount of time students at their college sleep
each night. Doreen and Jung each take samples of 500 students. Doreen
uses systematic sampling and Jung uses cluster sampling. Doreen's sample
will be different from Jung's sample. Even if Doreen and Jung used the

same sampling method, in all likelihood their samples would be different.
Neither would be wrong, however.

Think about what contributes to making Doreen's and Jung's samples
different.

If Doreen and Jung took larger samples (i.e. the number of data values is
increased), their sample results (the average amount of time a student
Sleeps) might be closer to the actual population average. But still, their
samples would be, in all likelihood, different from each other. This
variability in samples cannot be stressed enough.

Size of a Sample

The size of a sample (often called the number of observations) is important.
The examples you have seen in this book so far have been small. Samples
of only a few hundred observations, or even smaller, are sufficient for many
purposes. In polling, samples that are from 1200 to 1500 observations are
considered large enough and good enough if the survey is random and is
well done. You will learn why when you study confidence intervals.

Be aware that many large samples are biased. For example, call-in surveys
are invariable biased because people choose to respond or not.

Optional Collaborative Classroom Exercise

Exercise:

Problem:

Divide into groups of two, three, or four. Your instructor will give each
group one 6-sided die. Try this experiment twice. Roll one fair die (6-
sided) 20 times. Record the number of ones, twos, threes, fours, fives,
and sixes you get below ("frequency" is the number of times a
particular face of the die occurs):

Face on Die Frequency

i)

rs)
6

First Experiment (20 rolls)

Face on Die Frequency

i.

rs)
6

Second Experiment (20 rolls)

Did the two experiments have the same results? Probably not. If you
did the experiment a third time, do you expect the results to be
identical to the first or second experiment? (Answer yes or no.) Why
or why not?

Which experiment had the correct results? They both did. The job of
the statistician is to see through the variability and draw appropriate
conclusions.

Critical Evaluation

We need to critically evaluate the statistical studies we read about and
analyze before accepting the results of the study. Common problems to be
aware of include

e Problems with Samples: A sample should be representative of the
population. A sample that is not representative of the population is
biased. Biased samples that are not representative of the population
give results that are inaccurate and not valid.

e Self-Selected Samples: Responses only by people who choose to
respond, such as call-in surveys are often unreliable.

e Sample Size Issues: Samples that are too small may be unreliable.
Larger samples are better if possible. In some situations, small samples
are unavoidable and can still be used to draw conclusions, even though
larger samples are better. Examples: Crash testing cars, medical testing
for rare conditions.

e Undue influence: Collecting data or asking questions in a way that
influences the response.

e Non-response or refusal of subject to participate: The collected
responses may no longer be representative of the population. Often,
people with strong positive or negative opinions may answer surveys,
which can affect the results.

e Causality: A relationship between two variables does not mean that
one causes the other to occur. They may both be related (correlated)
because of their relationship through a different variable.

e Self-Funded or Self-Interest Studies: A study performed by a person or
organization in order to support their claim. Is the study impartial?

Read the study carefully to evaluate the work. Do not automatically
assume that the study is good but do not automatically assume the
study is bad either. Evaluate it on its merits and the work done.

e Misleading Use of Data: Improperly displayed graphs, incomplete
data, lack of context.

e Confounding: When the effects of multiple factors on a response
cannot be separated. Confounding makes it difficult or impossible to
draw valid conclusions about the effect of each factor.

Glossary

Population
The collection, or set, of all individuals, objects, or measurements
whose properties are being studied.

Sample
A portion of the population understudy. A sample is representative if it
characterizes the population being studied.

Answers and Rounding Off
This module briefly explains the correct way to round off answers when
working with statistical data.

A simple way to round off answers is to carry your final answer one more
decimal place than was present in the original data. Round only the final
answer. Do not round any intermediate results, if possible. If it becomes
necessary to round intermediate results, carry them to at least twice as many
decimal places as the final answer. For example, the average of the three
quiz scores 4, 6, 9 is 6.3, rounded to the nearest tenth, because the data are
whole numbers. Most answers will be rounded in this manner.

It is not necessary to reduce most fractions in this course. Especially in
Probability Topics, the chapter on probability, it is more helpful to leave an
answer as an unreduced fraction.

Frequency

This module introduces the concepts of frequency, relative frequency, and
cumulative relative frequency, and the relationship between these measures.
Students will have the opportunity to interpret data through the sample problems
provided.

Twenty students were asked how many hours they worked per day. Their
responses, in hours, are listed below:

96332475235654435253

Below is a frequency table listing the different data values in ascending order and
their frequencies.

DATA VALUE FREQUENCY
2 3
3 fs)
4 3
fs) 6
6 2
7 1

Frequency Table of Student Work Hours

A frequency is the number of times a given datum occurs in a data set.
According to the table above, there are three students who work 2 hours, five
students who work 3 hours, etc. The total of the frequency column, 20, represents
the total number of students included in the sample.

A relative frequency is the fraction or proportion of times an answer occurs. To
find the relative frequencies, divide each frequency by the total number of
students in the sample - in this case, 20. Relative frequencies can be written as
fractions, percents, or decimals.

DATA VALUE FREQUENCY RELATIVE FREQUENCY

2 3 +. or 0.15
3 5 2 or 0.25
A 3 x or 0.15
5 6 3; or 0.30
6 2 + or 0.10
7 1 $a or 0.05

Frequency Table of Student Work Hours w/ Relative Frequency

The sum of the relative frequency column is an or 1.

Cumulative relative frequency is the accumulation of the previous relative
frequencies. To find the cumulative relative frequencies, add all the previous
relative frequencies to the relative frequency for the current row.

CUMULATIVE

DATA RELATIVE RELATIVE
VALUE FREQUENCY FREQUENCY FREQUENCY
2 3 3 or 0.15 0.15

5 0.15 + 0.25 =
3 i) 59 OF 0.25 0.40

as 0.40 + 0.15 =
4 3 50 OF 0.15 0.55

6 0.55 + 0.30 =
5 6 59 OF 0.30 0.85

2 0.85 + 0.10 =
6 2 30 OF 0.10 0.95

ot) 0.95 + 0.05 =
7 1 59 OF 0.05 1.00

Frequency Table of Student Work Hours w/ Relative and Cumulative Relative
Frequency

The last entry of the cumulative relative frequency column is one, indicating that
one hundred percent of the data has been accumulated.

Note:Because of rounding, the relative frequency column may not always sum
to one and the last entry in the cumulative relative frequency column may not be
one. However, they each should be close to one.

The following table represents the heights, in inches, of a sample of 100 male
semiprofessional soccer players.

HEIGHTS
(INCHES)

59.95 -
61.95

61.95 -
63.95

63.95 -
65.95

65.95 -
67.95

67.95 -
69.95

69.95 -
71.95

71.95 -
7390

73.95 -
75.95

FREQUENCY

15

40

17

12

Total = 100

59.95 - 61.95 inches
61.95 - 63.95 inches
63.95 - 65.95 inches
65.95 - 67.95 inches
67.95 - 69.95 inches

RELATIVE
FREQUENCY
<a5 = 0.05

=35 = 0.03

G60 = 0-15

2. = 0.40

ay = 0.17

Fay = 0.12

aie = 0.07

500 = 0.01
Total = 1.00

Frequency Table of Soccer Player Height

CUMULATIVE
RELATIVE
FREQUENCY

0.05

0.05 + 0.03 =
0.08

0.08 + 0.15 =
0.23

0.23 + 0.40 =
0.63

0.63 + 0.17 =
0.80

0.80 + 0.12 =
0.92

0.92 + 0.07 =
0.99

0.99 + 0.01 =
1.00

The data in this table has been grouped into the following intervals:

e 69.95 - 71.95 inches
e 71.95 - 73.95 inches
e 73.95 - 75.95 inches

Note: This example is used again in the Descriptive Statistics chapter, where the
method used to compute the intervals will be explained.

In this sample, there are 5 players whose heights are between 59.95 - 61.95
inches, 3 players whose heights fall within the interval 61.95 - 63.95 inches, 15
players whose heights fall within the interval 63.95 - 65.95 inches, 40 players
whose heights fall within the interval 65.95 - 67.95 inches, 17 players whose
heights fall within the interval 67.95 - 69.95 inches, 12 players whose heights fall
within the interval 69.95 - 71.95, 7 players whose height falls within the interval
71.95 - 73.95, and 1 player whose height falls within the interval 73.95 - 75.95.
All heights fall between the endpoints of an interval and not at the endpoints.

Example:
Exercise:

Problem:

From the table, find the percentage of heights that are less than 65.95
inches.

Solution:

If you look at the first, second, and third rows, the heights are all less than
65.95 inches. There are 5 + 3 + 15 = 23 males whose heights are less than
65.95 inches. The percentage of heights less than 65.95 inches is then aan
or 23%. This percentage is the cumulative relative frequency entry in the

third row.

Example:

Exercise:

Problem:

From the table, find the percentage of heights that fall between 61.95 and
65.95 inches.

Solution:

Add the relative frequencies in the second and third rows: 0.03 + 0.15 =
0.18 or 18%.

Example:
Exercise:

Problem:

Use the table of heights of the 100 male semiprofessional soccer players.
Fill in the blanks and check your answers.

. The percentage of heights that are from 67.95 to 71.95 inches is:

. The percentage of heights that are from 67.95 to 73.95 inches is:

. The percentage of heights that are more than 65.95 inches is:

. The number of players in the sample who are between 61.95 and 71.95
inches tall is:

5. What kind of data are the heights?

6. Describe how you could gather this data (the heights) so that the data

are characteristic of all male semiprofessional soccer players.

BRWN Ee

Remember, you count frequencies. To find the relative frequency, divide
the frequency by the total number of data values. To find the cumulative
relative frequency, add all of the previous relative frequencies to the
relative frequency for the current row.

Solution:

led 76
2. 36%
Be 11 20

4. 87

5. quantitative continuous

6. get rosters from each team and choose a simple random sample from
each

Optional Collaborative Classroom Exercise

Exercise:

Problem:

In your class, have someone conduct a survey of the number of siblings
(brothers and sisters) each student has. Create a frequency table. Add to it a
relative frequency column and a cumulative relative frequency column.
Answer the following questions:

1. What percentage of the students in your class has 0 siblings?
2. What percentage of the students has from 1 to 3 siblings?
3. What percentage of the students has fewer than 3 siblings?

Example:

Nineteen people were asked how many miles, to the nearest mile they commute
to work each day. The data are as follows:

Is iene alerlsres0) vaalO list ays esuhowilerd Wise)

The following table was produced:

CUMULATIVE
RELATIVE RELATIVE
DATA FREQUENCY FREQUENCY FREQUENCY

CUMULATIVE
RELATIVE RELATIVE
DATA FREQUENCY FREQUENCY FREQUENCY

3 3 + 0.1579
4 i in 0.2105
5 3 4 0.1579
7 2 4 0.2632
10 3 = 0.4737
12 2 A 0.7895
13 1 ts 0.8421
15 1 a 0.8948
18 1 a. 0.9474
20 1 de 1.0000

Frequency of Commuting Distances

Exercise:

Problem:

1. Is the table correct? If it is not correct, what is wrong?

2. True or False: Three percent of the people surveyed commute 3 miles.
If the statement is not correct, what should it be? If the table is
incorrect, make the corrections.

3. What fraction of the people surveyed commute 5 or 7 miles?

4. What fraction of the people surveyed commute 12 miles or more?
Less than 12 miles? Between 5 and 13 miles (does not include 5 and
13 miles)?

Solution:

1. No. Frequency column sums to 18, not 19. Not all cumulative relative
frequencies are correct.

2. False. Frequency for 3 miles should be 1; for 2 miles (left out), 2.
Cumulative relative frequency column should read: 0.1052, 0.1579,
0.2105, 0.3684, 0.4737, 0.6316, 0.7368, 0.7895, 0.8421, 0.9474, 1.

5
By ioeemre
4.39 19> Ig
Glossary
Frequency

The number of times a value of the data occurs.

Relative Frequency

The ratio of the number of times a value of the data occurs in the set of all
outcomes to the number of all outcomes.

Cumulative Relative Frequency

The term applies to an ordered set of observations from smallest to largest.
The Cumulative Relative Frequency is the sum of the relative frequencies
for all values that are less than or equal to the given value.

Summary

This module provides an outline/review of key concepts related to statistical
sampling and data.

Statistics

¢ Deals with the collection, analysis, interpretation, and presentation of
data

Probability
e Mathematical tool used to study randomness
Key Terms

e Population
e Parameter
e Sample

e Statistic

e Variable

e Data

Types of Data
¢ Quantitative Data (a number)

o Discrete (You count it.)
o Continuous (You measure it.)

¢ Qualitative Data (a category, words)
Sampling
¢ With Replacement: A member of the population may be chosen more
than once
e Without Replacement: A member of the population may be chosen

only once

Random Sampling

e Each member of the population has an equal chance of being selected
Sampling Methods
e Random

Simple random sample
Stratified sample
Cluster sample
Systematic sample

oO 0 0

e Not Random
o Convenience sample
Frequency (freq. or f)
e The number of times an answer occurs
Relative Frequency (rel. freq. or RF)

e The proportion of times an answer occurs
e Can be interpreted as a fraction, decimal, or percent

Cumulative Relative Frequencies (cum. rel. freq. or cum RF)

e An accumulation of the previous relative frequencies

Practice: Sampling and Data

This module provides an opportunity for students to practice concepts
related to statistical sampling and data. Given a sample data set, the student
will practice constructing frequency tables, differentiating between key
terms, and comparing sampling techniques.

Student Learning Outcomes

e The student will construct frequency tables.
e The student will differentiate between key terms.
e The student will compare sampling techniques.

Given

Studies are often done by pharmaceutical companies to determine the
effectiveness of a treatment program. Suppose that a new AIDS antibody
drug is currently under study. It is given to patients once the AIDS
symptoms have revealed themselves. Of interest is the average(mean)
length of time in months patients live once starting the treatment. Two
researchers each follow a different set of 40 AIDS patients from the start of
treatment until their deaths. The following data (in months) are collected.

Researcher A3 4 11 15 16 17 22 44 37 16 14 24 25 15 26 27 33 29 35 44
13 21 22 10 12 8 40 32 26 27 31 34 29 17 8 24 18 47 33 34

Researcher B3 14 115 16 17 28 41 31 18 14 14 26 25 21 22 31 2 35 44 23
2121 16 12 18 41 22 16 25 33 34 29 13 18 24 23 42 33 29

Organize the Data

Complete the tables below using the data provided.

Cumulative
Survival Length Relative Relative
(in months) Frequency Frequency Frequency
0.5-6.5
6.5-12.5
12.5 - 18.5
18.5 - 24.5
24.5 - 30.5
30.5 - 36.5
36.5 - 42.5

42.5 - 48.5

Researcher A

Cumulative
Survival Length Relative Relative
(in months) Frequency Frequency Frequency
0.5-6.5
6.5-12.5

12.5 - 18.5

Cumulative

Survival Length Relative Relative

(in months) Frequency Frequency Frequency
18.5 - 24.5

24.5 - 30.5

30.5 - 36.5

36.5 - 42.5

42.5 - 48.5

Researcher B

Key Terms

Define the key terms based upon the above example for Researcher A.
Exercise:

Problem: Population

Exercise:

Problem: Sample

Exercise:

Problem: Parameter

Exercise:

Problem: Statistic

Exercise:

Problem: Variable

Exercise:

Problem: Data

Discussion Questions

Discuss the following questions and then answer in complete sentences.
Exercise:

Problem: List two reasons why the data may differ.
Exercise:
Problem:
Can you tell if one researcher is correct and the other one is incorrect?
Why?

Exercise:

Problem: Would you expect the data to be identical? Why or why not?

Exercise:

Problem: How could the researchers gather random data?
Exercise:
Problem:
Suppose that the first researcher conducted his survey by randomly
choosing one state in the nation and then randomly picking 40 patients

from that state. What sampling method would that researcher have
used?

Exercise:

Problem:

Suppose that the second researcher conducted his survey by choosing
AO patients he knew. What sampling method would that researcher
have used? What concerns would you have about this data set, based
upon the data collection method?

Homework

This module presents students with a number of problems related to
Statistical sampling and data. In particular, students are asked to
demonstrate understanding of concepts such as frequency, relative
frequency, and cumulative relative frequency, random samples, quantitative
vs. qualitative data, continuous vs. discrete data, and other key terms related
to sampling and data.

Exercise:

Problem: For each item below:

e ildentify the type of data (quantitative - discrete, quantitative -
continuous, or qualitative) that would be used to describe a
response.

e iiGive an example of the data.

e aNumber of tickets sold to a concert

¢ bAmount of body fat

e cFavorite baseball team

e dTime in line to buy groceries

e eNumber of students enrolled at Evergreen Valley College

e fMost—watched television show

e gBrand of toothpaste

e hDistance to the closest movie theatre

e iAge of executives in Fortune 500 companies

e jNumber of competing computer spreadsheet software packages

Solution:

e aquantitative - discrete

e bquantitative - continuous
¢ cqualitative

e dquantitative - continuous
e equantitative - discrete

e fqualitative

¢ gqualitative

e hquantitative - continuous
e iquantitative - continuous
e jquantitative - discrete

Exercise:

Problem:

Fifty part-time students were asked how many courses they were
taking this term. The (incomplete) results are shown below:

# of
Courses

3

Relative
Frequency Frequency
30 0.6

15

Part-time Student Course Loads

e aFill in the blanks in the table above.
e bWhat percent of students take exactly two courses?
e¢ cWhat percent of students take one or two courses?

Exercise:

Cumulative
Relative
Frequency

Problem:

Sixty adults with gum disease were asked the number of times per
week they used to floss before their diagnoses. The (incomplete)
results are shown below:

# Flossing

per Week Frequency
0 27

1 18

3

6 3

7 ii

Relative
Frequency

0.4500

0.0500

0.0167

Flossing Frequency for Adults with Gum Disease

e aFill in the blanks in the table above.
e bWhat percent of adults flossed six times per week?
¢ cWhat percent flossed at most three times per week?

Solution:

e aCum. Rel. Freq. for 0 is 0.4500

Cumulative
Relative
Freq.

0.9333

Rel. Freq. for 1 is 0.3000 and Cum. Rel. Freq. for 1 or less is

0.7500

Freq. for 3 is 11 and Rel. Freq. is 0.1833
Cum. Rel. Freq. for 6 or less is 0.9833
Cum. Rel. Freq. for 7 or less is 1

¢ b5.00%

© €93.33%

Exercise:

Problem:

A fitness center is interested in the mean amount of time a client
exercises in the center each week. Define the following in terms of the
study. Give examples where appropriate.

e aPopulation
e bSample

e cParameter
e dStatistic

e eVariable

e fData

Exercise:

Problem:

Ski resorts are interested in the mean age that children take their first
ski and snowboard lessons. They need this information to optimally
plan their ski classes. Define the following in terms of the study. Give
examples where appropriate.

e aPopulation
e bSample

e cParameter
e dStatistic

e eVariable

e fData

Solution:

e aChildren who take ski or snowboard lessons

e bA group of these children

e cThe population mean

e dThe sample mean

e eX = the age of one child who takes the first ski or snowboard
lesson

e fValues for X, such as 3, 7, etc.

Exercise:

Problem:

A cardiologist is interested in the mean recovery period for her patients
who have had heart attacks. Define the following in terms of the study.
Give examples where appropriate.

e aPopulation
e bSample

e cParameter
e dStatistic

e eVariable

e fData

Exercise:

Problem:

Insurance companies are interested in the mean health costs each year
for their clients, so that they can determine the costs of health
insurance. Define the following in terms of the study. Give examples
where appropriate.

e aPopulation
e bSample
e cParameter
e dStatistic

e eVariable
e fData

Solution:

e aThe clients of the insurance companies
e bA group of the clients

e cThe mean health costs of the clients

e dThe mean health costs of the sample

e eX = the health costs of one client

e fValues for X, such as 34, 9, 82, etc.

Exercise:

Problem:

A politician is interested in the proportion of voters in his district that
think he is doing a good job. Define the following in terms of the
study. Give examples where appropriate.

e aPopulation
e bSample

e cParameter
e dStatistic

e eVariable

e fData

Exercise:
Problem:
A marriage counselor is interested in the proportion the clients she

counsels that stay married. Define the following in terms of the study.
Give examples where appropriate.

e aPopulation
e bSample

e cParameter
e dStatistic
e eVariable
e fData

Solution:

e aAll the clients of the counselor

¢ bA group of the clients

e cThe proportion of all her clients who stay married
e dThe proportion of the sample who stay married

e eX =the number of couples who stay married

e fyes, no

Exercise:

Problem:

Political pollsters may be interested in the proportion of people that
will vote for a particular cause. Define the following in terms of the
study. Give examples where appropriate.

e aPopulation
e bSample

e cParameter
e dStatistic

e eVariable

e fData

Exercise:
Problem:
A marketing company is interested in the proportion of people that will

buy a particular product. Define the following in terms of the study.
Give examples where appropriate.

e aPopulation
e bSample

e cParameter
e dStatistic

e eVariable

e fData

Solution:

¢ aAll people (maybe in a certain geographic area, such as the
United States)

e bA group of the people

e cThe proportion of all people who will buy the product

e dThe proportion of the sample who will buy the product

e eX =the number of people who will buy it

e fbuy, not buy

Exercise:

Problem:

Airline companies are interested in the consistency of the number of
babies on each flight, so that they have adequate safety equipment.
Suppose an airline conducts a survey. Over Thanksgiving weekend, it
surveys 6 flights from Boston to Salt Lake City to determine the
number of babies on the flights. It determines the amount of safety
equipment needed by the result of that study.

e aUsing complete sentences, list three things wrong with the way
the survey was conducted.

e bUsing complete sentences, list three ways that you would
improve the survey if it were to be repeated.

Exercise:

Problem:

Suppose you want to determine the mean number of students per
Statistics class in your state. Describe a possible sampling method in 3
— 5 complete sentences. Make the description detailed.

Exercise:
Problem:
Suppose you want to determine the mean number of cans of soda
drunk each month by persons in their twenties. Describe a possible

sampling method in 3 - 5 complete sentences. Make the description
detailed.

Exercise:
Problem:
771 distance learning students at Long Beach City College responded
to surveys in the 2010-11 academic year. Highlights of the summary

report are listed in the table below. (Source:
http://de.lbcc.edu/reports/2010-11/future/highlights.html#focus).

Have computer at home 96%
Unable to come to campus for classes 65%
Age 41 or over 24%
Would like LBCC to offer more DL courses 95%
Took DL classes due to a disability 17%

Live at least 16 miles from campus 13%

Took DL courses to fulfill transfer requirements 71%
LBCC Distance Learning Survey Results

e¢ aWhat percent of the students surveyed do not have a computer at
home?

e bAbout how many students in the survey live at least 16 miles
from campus?

e clf the same survey was done at Great Basin College in Elko,
Nevada, do you think the percentages would be the same? Why?

Solution:

e a4%
e b100

Exercise:
Problem:

Nineteen immigrants to the U.S were asked how many years, to the
nearest year, they have lived in the U.S. The data are as follows:

20/22102015:070.20512151245.10

The following table was produced:

Relative Cumulative
Data Frequency Frequency Relative Frequency

0 2 4 0.1053

Relative Cumulative

Data Frequency Frequency Relative Frequency
2 3 a 0.2632
4 5 0.3158
5 3 y 0.1579
7 2 45 0.5789
10 2 4 0.6842
12 2 4 0.7895
15 1 5 0.8421
20 1 + 1.0000

Frequency of Immigrant Survey Responses

aFix the errors on the table. Also, explain how someone might
have arrived at the incorrect number(s).

bExplain what is wrong with this statement: “47 percent of the
people surveyed have lived in the U.S. for 5 years.”

cFix the statement above to make it correct.

dWhat fraction of the people surveyed have lived in the U.S. 5 or
7 years?

eWhat fraction of the people surveyed have lived in the U.S. at
most 12 years?

fWhat fraction of the people surveyed have lived in the U.S.
fewer than 12 years?

gWhat fraction of the people surveyed have lived in the U.S. from
5 to 20 years, inclusive?

Exercise:

Problem:

A “random survey” was conducted of 3274 people of the
“microprocessor generation” (people born since 1971, the year the
microprocessor was invented). It was reported that 48% of those
individuals surveyed stated that if they had $2000 to spend, they would
use it for computer equipment. Also, 66% of those surveyed
considered themselves relatively savvy computer users. (Source: San
Jose Mercury News)

e aDo you consider the sample size large enough for a study of this
type? Why or why not?

e bBased on your “gut feeling,” do you believe the percents
accurately reflect the U.S. population for those individuals born
since 1971? If not, do you think the percents of the population are
actually higher or lower than the sample statistics? Why?

Additional information: The survey was reported by Intel Corporation
of individuals who visited the Los Angeles Convention Center to see
the Smithsonian Institure's road show called “America’s Smithsonian.”

e cWith this additional information, do you feel that all
demographic and ethnic groups were equally represented at the
event? Why or why not?

e dWith the additional information, comment on how accurately
you think the sample statistics reflect the population parameters.

Exercise:

Problem:

e aList some practical difficulties involved in getting accurate
results from a telephone survey.

e bList some practical difficulties involved in getting accurate
results from a mailed survey.

e cWith your classmates, brainstorm some ways to overcome these
problems if you needed to conduct a phone or mail survey.

Try these multiple choice questions

The next four questions refer to the following: A Lake Tahoe
Community College instructor is interested in the mean number of days
Lake Tahoe Community College math students are absent from class during
a quarter.

Exercise:

Problem: What is the population she is interested in?

e AAIl Lake Tahoe Community College students

e BAI] Lake Tahoe Community College English students

e CAIl Lake Tahoe Community College students in her classes
e DAI Lake Tahoe Community College math students

Solution:

D

Exercise:

Problem: Consider the following:

X = number of days a Lake Tahoe Community College math
student is absent

In this case, X is an example of a:

e AVariable
e BPopulation
e CStatistic

e DData

Solution:

A
Exercise:
Problem:
The instructor takes her sample by gathering data on 5 randomly

selected students from each Lake Tahoe Community College math
class. The type of sampling she used is

e ACluster sampling

e BStratified sampling

e CSimple random sampling
e DConvenience sampling

Solution:

B
Exercise:

Problem:

The instructor’s sample produces an mean number of days absent of
3.5 days. This value is an example of a

e AParameter
e BData

e CStatistic
e DVariable

Solution:

G

The next two questions refer to the following relative frequency table on
hurricanes that have made direct hits on the U.S between 1851 and 2004.
Hurricanes are given a strength category rating based on the minimum wind
speed generated by the storm. (http://www.nhc.noaa. gov/gifs/table5. gif)

Category
1
Z

3

Frequency of Hurricane Direct Hits

Exercise:

Problem:

Number of
Direct Hits

109

72

71

18

3

Total = 273

Relative
Frequency

0.3993
0.2637

0.2601

0.0110

Cumulative
Frequency

0.3993

0.6630

0.9890

1.0000

What is the relative frequency of direct hits that were category 4

hurricanes?

e A0.0768
e B0.0659
e C0.2601

e DNot enough information to calculate

Solution:

B
Exercise:

Problem:

What is the relative frequency of direct hits that were AT MOST a
category 3 storm?

e A0.3480
e B0.9231
e C0.2601
¢ DO0.3370

Solution:

B

The next three questions refer to the following: A study was done to
determine the age, number of times per week and the duration (amount of
time) of resident use of a local park in San Jose. The first house in the
neighborhood around the park was selected randomly and then every 8th
house in the neighborhood around the park was interviewed.

Exercise:

Problem: "‘Number of times per week’" is what type of data?
e Aqualitative

e Bquantitative - discrete
e Cquantitative - continuous

Solution:

B

Exercise:

Problem: The sampling method was:

e Asimple random
e Bsystematic

¢ Cstratified

¢ Dcluster

Solution:

B

Exercise:

ou

Problem: "‘Duration (amount of time)’" is what type of data?

e Aqualitative
e Bquantitative - discrete
e Cquantitative - continuous

Solution:

C

Exercises 28 and 29 are not multiple choice exercises.
Exercise:

Problem:
Name the sampling method used in each of the following situations:

e AA woman in the airport is handing out questionnaires to
travelers asking them to evaluate the airport’s service. She does
not ask travelers who are hurrying through the airport with their

hands full of luggage, but instead asks all travelers sitting near
gates and who are not taking naps while they wait.

BA teacher wants to know if her students are doing homework so
she randomly selects rows 2 and 5, and then calls on all students
in row 2 and all students in row 5 to present the solution to
homework problems to the class.

CThe marketing manager for an electronics chain store wants
information about the ages of its customers. Over the next two
weeks, at each store location, 100 randomly selected customers
are given questionnaires to fill out which asks for information
about age, as well as about other variables of interest.

DThe librarian at a public library wants to determine what
proportion of the library users are children. The librarian has a
tally sheet on which she marks whether the books are checked out
by an adult or a child. She records this data for every 4th patron
who checks out books.

EA political party wants to know the reaction of voters to a
debate between the candidates. The day after the debate, the
party’s polling staff calls 1200 randomly selected phone numbers.
If a registered voter answers the phone or is available to come to
the phone, that registered voter is asked who he/she intends to
vote for and whether the debate changed his/her opinion of the
candidates.

** Contributed by Roberta Bloom

Solution:

AConvenience
BCluster
CStratified
DSystematic
ESimple Random

Exercise:

Problem:

Several online textbook retailers advertise that they have lower prices
than on-campus bookstores. However, an important factor is whether
the internet retailers actually have the textbooks that students need in
stock. Students need to be able to get textbooks promptly at the
beginning of the college term. If the book is not available, then a
student would not be able to get the textbook at all, or might get a
delayed delivery if the book is back ordered.

A college newspaper reporter is investigating textbook availability at
online retailers. He decides to investigate one textbook for each of the
following 7 subjects: calculus, biology, chemistry, physics, statistics,
geology, and general engineering. He consults textbook industry sales
data and selects the most popular nationally used textbook in each of
these subjects. He visits websites for a random sample of major online
textbook sellers and looks up each of these 7 textbooks to see if they
are available in stock for quick delivery through these retailers. Based
on his investigation, he writes an article in which he draws conclusions
about the overall availability of all college textbooks through online
textbook retailers.

Write an analysis of his study that addresses the following issues: Is
his sample representative of the population of all college textbooks?
Explain why or why not. Describe some possible sources of bias in this
study, and how it might affect the results of the study. Give some
suggestions about what could be done to improve the study.

** Contributed by Roberta Bloom

Solution:

The answer below contains some of the issues that students might
discuss for this problem. Individual student's answers may also

identify other issues that pertain to this problem that are not included
in the answer below.

The sample is not representative of the population of all college
textbooks. Two reasons why it is not representative are that he only
sampled 7 subjects and he only investigated one textbook in each
subject. There are several possible sources of bias in the study. The 7
subjects that he investigated are all in mathematics and the sciences;
there are many subjects in the humanities, social sciences, and many
other subject areas, (for example: literature, art, history, psychology,
sociology, business) that he did not investigate at all. It may be that
different subject areas exhibit different patterns of textbook
availability, but his sample would not detect such results.

He also only looked at the most popular textbook in each of the
subjects he investigated. The availability of the most popular textbooks
may differ from the availability of other textbooks in one of two ways:

e the most popular textbooks may be more readily available online,
because more new copies are printed and more students
nationwide selling back their used copies OR

e the most popular textbooks may be harder to find available
online, because more student demand exhausts the supply more
quickly.

In reality, many college students do not use the most popular textbook
in their subject, and this study gives no useful information about the
situation for those less popular textbooks.

He could improve this study by

e expanding the selection of subjects he investigates so that it is
more representative of all subjects studied by college students and

e expanding the selection of textbooks he investigates within each
subject to include a mixed representation of both the popular and
less popular textbooks.

Introduction

Student Learning Outcomes
By the end of this chapter, the student should be able to:

e Display data graphically and interpret graphs: stemplots, histograms
and boxplots.

e Recognize, describe, and calculate the measures of location of data:
quartiles and percentiles.

e Recognize, describe, and calculate the measures of the center of data:
mean, median, and mode.

e Recognize, describe, and calculate the measures of the spread of data:
variance, standard deviation, and range.

Introduction

Once you have collected data, what will you do with it? Data can be
described and presented in many different formats. For example, suppose
you are interested in buying a house in a particular area. You may have no
clue about the house prices, so you might ask your real estate agent to give
you a sample data set of prices. Looking at all the prices in the sample often
is overwhelming. A better way might be to look at the median price and the
variation of prices. The median and variation are just two ways that you
will learn to describe data. Your agent might also provide you with a graph
of the data.

In this chapter, you will study numerical and graphical ways to describe and
display your data. This area of statistics is called "Descriptive Statistics".
You will learn to calculate, and even more importantly, to interpret these
measurements and graphs.

Displaying Data
This module provides a brief introduction into the ways graphs and charts
can be used to provide visual representations of data.

A Statistical graph is a tool that helps you learn about the shape or
distribution of a sample. The graph can be a more effective way of
presenting data than a mass of numbers because we can see where data
clusters and where there are only a few data values. Newspapers and the
Internet use graphs to show trends and to enable readers to compare facts
and figures quickly.

Statisticians often graph data first to get a picture of the data. Then, more
formal tools may be applied.

Some of the types of graphs that are used to summarize and organize data
are the dot plot, the bar chart, the histogram, the stem-and-leaf plot, the
frequency polygon (a type of broken line graph), pie charts, and the
boxplot. In this chapter, we will briefly look at stem-and-leaf plots, line
graphs and bar graphs. Our emphasis will be on histograms and boxplots.

Graphing Qualitative Variables

When Apple Computer introduced the iMac computer in August 1998, the
company wanted to learn whether the iMac was expanding Apple?s market
share. Was the iMac just attracting previous Macintosh owners? Or was it
purchased by newcomers to the computer market, and by previous
Windows users who were switching over? To find out, 500 iMac customers
were interviewed. Each customer was categorized as a previous Macintosh
Owners, a previous Windows owner, or a new computer purchaser. This
section examines graphical methods for displaying the results of the
interviews. We'll learn some general lessons about how to graph data that
fall into a small number of categories. A later section will consider how to
graph numerical data in which each observation is represented by a number
in some range. The key point about the qualitative data that occupy us in the
present section is that they do not come with a pre-established ordering (the
way numbers are ordered). For example, there is no natural sense in which
the category of previous Windows users comes before or after the category
of previous iMac users. This situation may be contrasted with quantitative
data, such as a person?s weight. People of one weight are naturally ordered
with respect to people of a different weight.

Frequency Tables

All of the graphical methods shown in this section are derived from
frequency tables. [link] shows a frequency table for the results of the iMac
study; it shows the frequencies of the various response categories. It also
shows the relative frequencies, which are the proportion of responses in
each category. For example, the relative frequency for "none" of

0.17 = 85/500.

Previous Ownership Frequency Relative Frequency

Previous Ownership
None

Windows

Macintosh

Total

Frequency
85

60

355

500

Frequency Table for the Mac Data

Pie Charts

Relative Frequency
0.17
0.12
0.71

1.00

The pie chart in [link] shows the results of the iMac study. In a pie chart,
each category is represented by a slice of the pie. The area of the slice is
proportional to the percentage of responses in the category. This is simply
the relative frequency multiplied by 100. Although most iMac purchasers
were Macintosh owners, Apple was encouraged by the 12% of purchasers
who were former Windows users, and by the 17% of purchasers who were
buying a computer for the first time.

None
17%

Windows
12%

Macintosh
71%

Pie chart of iMac

purchases illustrating

frequencies of

previous computer
ownership.

Pie charts are effective for displaying the relative frequencies of a small
number of categories. They are not recommended, however, when you have
a large number of categories. Pie charts can also be confusing when they are
used to compare the outcomes of two different surveys or experiments. In
an influential book on the use of graphs, Edward Tufte asserted ""The only

LARA

worse design than a pie chart is several of them"".

Here is another important point about pie charts. If they are based on a
small number of observations, it can be misleading to label the pie slices
with percentages. For example, if just 5 people had been interviewed by
Apple Computers, and 3 were former Windows users, it would be
misleading to display a pie chart with the Windows slice showing 60%.
With so few people interviewed, such a large percentage of Windows users
might easily have accord since chance can cause large errors with small
samples. In this case, it is better to alert the user of the pie chart to the
actual numbers involved. The slices should therefore be labeled with the
actual frequencies observed (e.g., 3) instead of with percentages.

Bar Charts

Bar charts can also be used to represent frequencies of different categories.
A bar chart of the iMac purchases is shown in [link]. Frequencies are shown
on the Y axis and the type of computer previously owned is shown on the X
axis. Typically the Y-axis shows the number of observations rather than the
percentage of observations in each category as is typical in pie charts.

Number of Buyers
Nm
o
oe

None Windows Macintosh
Previous Computer

Bar chart of iMac purchases as a function
of previous computer ownership.

Note:EJA guidelines for statistical graphs.

Comparing Distributions

Often we need to compare the results of different surveys, or of different
conditions within the same overall survey. In this case, we are comparing
the distributions of responses between the surveys or conditions. Bar
charts are often excellent for illustrating differences between two
distributions. [link] shows the number of people playing card games at the
Yahoo website on a Sunday and on a Wednesday on a day in the Spring of
2001. We see that there were more players overall on Wednesday compared
to Sunday. The number of people playing Pinochle was nonetheless the
same on these two days. In contrast, there were about twice as many people
playing hearts on Wednesday as on Sunday. Facts like these emerge clearly
from a well-designed bar chart.

Spades
Euchre
Pinochle
Canasta
eal @ Sunday
Cribbage pea Wednesday
Gin
Bridge

Black jack

Poker

0 2000 4000 6000 8000

A bar chart of the number of people playing
different card games on Sunday and Wednesday.

The bars in [link] are oriented horizontally rather than vertically. The
horizontal format is useful when you have many categories because there is
more room for the category labels. Such horizontal bar charts may be
contrasted with vertical bar charts like the one in [link].

Note: Another example using bar charts to compare distributions.

We'll have more to say about bar charts when we consider numerical
quantities later in this chapter. (See Bar Charts.)

Some graphical mistakes to avoid

Don't get fancy! People sometimes add features to graphs that don't help to
convey their information. For example, 3-dimensional bar charts like the

one shown in [link] are usually not as effective as their two-dimensional
counterparts.

Windows

Macintosh

A three-dimensional
version of [link].

Here is another way that fanciness can lead to trouble. Instead of plain bars,
it is tempting to substitute meaningful images. For example, [link] presents
the iMac data using pictures of computers. The heights of the pictures
accurately represent the number of buyers, yet [link] is misleading because
the viewer's attention will be captured by areas. This can exaggerate the size
differences between the groups. In terms of percentages, the ratio of
previous Macintosh owners to previous Windows owners is about 6 to 1.
But the ratio of the two areas in [link] is about 35 to 1. A biased person
wishing to hide the fact that many Windows owners purchased iMacs
would be tempted to use [link] instead of [link]! Edward Tufte coined the
term lie factor to refer to the ratio of the size of the effect shown in a graph
to the size of the effect shown in the data. He suggests that lie factors
greater than 1.05 or less than 0.95 produce unacceptable distortion.

Number of Buyers

None Vindows Macintosh

Previous Computer

A redrawing of [link] with a lie
factor greater than 8.

Another distortion in bar charts results from setting the baseline to a value
other than zero. The baseline is the bottom of the Y-axis, representing the
least number of cases that could have occurred in a category. Normally, this
number should be zero. [link] shows the iMac data with a baseline of 50.
Once again, the difference in areas suggests a different story than the true
differences in percentages. The number of Windows-switchers seems
minuscule compared to its true value of 12%.

400

tw
ol
Oo

"
he

300

NM
on
Oo

ow
o oOo

Number of Buye
Nm
Qo
o

on
Oo

None Windows Macintosh
Previous Computer

A redrawing of [link] with a baseline of
50.

Finally, we note that it is a serious mistake to use a line graph when the X-
axis contains merely qualitative variables. A line graph is essentially a bar
graph with the tops of the bars represented by points joined by lines (the
rest of the bar is suppressed). [link] inappropriately shows a line graph of
the card game data from Yahoo. The drawback to [link] is that it gives the
false impression that the games are naturally ordered in a numerical way.

| |—@— Wednesday
—m— Sunday

a
a
x
[=]
a

Black jack
Bridge
Gin
Cribbage
Hearts
Canasta
Pinochle
Euchre
Spades

A line graph of the number of people playing
different card games on Sunday and Wednesday.

Summary

Pie charts and bar charts can both be effective methods of portraying
qualitative data. Bar charts are better when there are more than just a few
categories and for comparing two or more distributions. Be careful to avoid
creating misleading graphs.

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Histograms

This module provides an overview of Descriptive Statistics: Histogram as a
part of Collaborative Statistics collection (col10522) by Barbara Illowsky
and Susan Dean.

For most of the work you do in this book, you will use a histogram to
display the data. One advantage of a histogram is that it can readily display
large data sets. A rule of thumb is to use a histogram when the data set
consists of 100 values or more.

A histogram consists of contiguous boxes. It has both a horizontal axis and
a vertical axis. The horizontal axis is labeled with what the data represents
(for instance, distance from your home to school). The vertical axis is
labeled either Frequency or relative frequency. The graph will have the
same shape with either label. The histogram (like the stemplot) can give
you the shape of the data, the center, and the spread of the data. (The next
section tells you how to calculate the center and the spread.)

The relative frequency is equal to the frequency for an observed value of
the data divided by the total number of data values in the sample. (In the
chapter on Sampling and Data, we defined frequency as the number of
times an answer occurs.) If:

e f = frequency

e mn = total number of data values (or the sum of the individual
frequencies), and

¢ RF = relative frequency,

then:
Equation:

RF = 2
n

For example, if 3 students in Mr. Ahab's English class of 40 students
received from 90% to 100%, then,

a = Gol tees Cola ce.
f=3,n=40,andRF= — Go = 0.075

Seven and a half percent of the students received 90% to 100%. Ninety
percent to 100 % are quantitative measures.

To construct a histogram, first decide how many bars or intervals, also
called classes, represent the data. Many histograms consist of from 5 to 15
bars or classes for clarity. Choose a starting point for the first interval to be
less than the smallest data value. A convenient starting point is a lower
value carried out to one more decimal place than the value with the most
decimal places. For example, if the value with the most decimal places is
6.1 and this is the smallest value, a convenient starting point is 6.05 (6.1 -
0.05 = 6.05). We say that 6.05 has more precision. If the value with the
most decimal places is 2.23 and the lowest value is 1.5, a convenient
Starting point is 1.495 (1.5 - 0.005 = 1.495). If the value with the most
decimal places is 3.234 and the lowest value is 1.0, a convenient starting
point is 0.9995 (1.0 - .0005 = 0.9995). If all the data happen to be integers
and the smallest value is 2, then a convenient starting point is 1.5 (2 - 0.5 =
1.5). Also, when the starting point and other boundaries are carried to one
additional decimal place, no data value will fall on a boundary.

Example:

The following data are the heights (in inches to the nearest half inch) of
100 male semiprofessional soccer players. The heights are continuous data
since height is measured.

60 60.5 61 61 61.5

63.5 63.5 63.5

64 64 64 64 64 64 64 64.5 64.5 64.5 64.5 64.5 64.5 64.5 64.5

66 66 66 66 66 66 66 66 66 66 66.5 66.5 66.5 66.5 66.5 66.5 66.5 66.5
66:5°66.5.56.0.67 67 67 67 67 67 67.67 67 67 67 57 67.5679 67.5675
OO Osc 0723

66-08 69:69 16959 69 69'69'69 6969. 69.5 69.5-69.9 0015 Gor

7070-70 7070 70;70.5: 7059 70.9 71 71 7A

T2722 72 2 oO 3-8

74

The smallest data value is 60. Since the data with the most decimal places
has one decimal (for instance, 61.5), we want our starting point to have two
decimal places. Since the numbers 0.5, 0.05, 0.005, etc. are convenient
numbers, use 0.05 and subtract it from 60, the smallest value, for the
convenient starting point.

60 - 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal
place. The starting point is, then, 59.95.

The largest value is 74. 74+ 0.05 = 74.05 is the ending value.

Next, calculate the width of each bar or class interval. To calculate this
width, subtract the starting point from the ending value and divide by the
number of bars (you must choose the number of bars you desire). Suppose
you choose 8 bars.

Equation:

74.05 — 59.95

— eG
8

Note: We will round up to 2 and make each bar or class interval 2 units
wide. Rounding up to 2 is one way to prevent a value from falling on a
boundary. Rounding to the next number is necessary even if it goes
against the standard rules of rounding. For this example, using 1.76 as the
width would also work.

The boundaries are:

* 59.95

Li trayw elo sae nee on Less)
Lda ee eae ak oo} 9/8)
S26) Joe too
© Ga oee 7195
LAV rehe sae eee fa)
eeOo ore = liao
e 71.95 + 2 = 73.95

73952 = / aos

The heights 60 through 61.5 inches are in the interval 59.95 - 61.95. The
heights that are 63.5 are in the interval 61.95 - 63.95. The heights that are
64 through 64.5 are in the interval 63.95 - 65.95. The heights 66 through
67.5 are in the interval 65.95 - 67.95. The heights 68 through 69.5 are in
the interval 67.95 - 69.95. The heights 70 through 71 are in the interval
69.95 - 71.95. The heights 72 through 73.5 are in the interval 71.95 -
73.95. The height 74 is in the interval 73.95 - 75.95.

The following histogram displays the heights on the x-axis and relative
frequency on the y-axis.

Relative
Frequency

04

59.95 61.95 63.95 65.95 67.95 69.95 71.95 73.95 75.95

Heights

Example:
The following data are the number of books bought by 50 part-time college
students at ABC College. The number of books is discrete data since books

are counted.

SoD ene Reale Aly hela

Dee ee

Sicushenershe mel oeomelogele roi,

444444

Deore

66

Eleven students buy 1 book. Ten students buy 2 books. Sixteen students
buy 3 books. Six students buy 4 books. Five students buy 5 books. Two
students buy 6 books.

Because the data are integers, subtract 0.5 from 1, the smallest data value
and add 0.5 to 6, the largest data value. Then the starting point is 0.5 and
the ending value is 6.5.

Exercise:

Problem:

Next, calculate the width of each bar or class interval. If the data are
discrete and there are not too many different values, a width that
places the data values in the middle of the bar or class interval is the
most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6
and the starting point is 0.5, a width of one places the 1 in the middle
of the interval from 0.5 to 1.5, the 2 in the middle of the interval from
1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in
the middle of the interval from to , the 5 in the
middle of the interval from to , and the in
the middle of the interval from to

Solution:

e 3.5t0 4.5
e 4.5t05.5
° 6

e 5.5 to 6.5

Calculate the number of bars as follows:
Equation:

6.5—0.5 _ 1
bars

where 1 is the width of a bar. Therefore, bars = 6.
The following histogram displays the number of books on the x-axis and

the frequency on the y-axis.

16
14__
12__
1
Py
3
=
g
i
2
0.5 1.5 2.5 a5 4.5 3 6.5

Number of Books

Using the TI-83, 83+, 84, 84+ Calculator Instructions

Go to the Appendix (14:Appendix) in the menu on the left. There are
calculator instructions for entering data and for creating a customized
histogram. Create the histogram for Example 2.

e Press Y=. Press CLEAR to clear out any equations.

e Press STAT 1:EDIT. If L1 has data in it, arrow up into the name L1,
press CLEAR and arrow down. If necessary, do the same for L2.

e Into L1, enter 1, 2, 3, 4, 5,6

e Into L2, enter 11, 10, 16, 6, 5, 2

e Press WINDOW. Make Xmin = .5, Xmax = 6.5, Xscl = (6.5 - .5)/6,
Ymin = -1, Ymax = 20, Yscl = 1, Xres = 1

e Press 2nd Y=. Start by pressing 4:Plotsoff ENTER.

e Press 2nd Y=. Press 1:Plot1. Press ENTER. Arrow down to TYPE.
Arrow to the 3rd picture (histogram). Press ENTER.

e Arrow down to Xlist: Enter L1 (2nd 1). Arrow down to Freq. Enter L2
(2nd 2).

e Press GRAPH

e Use the TRACE key and the arrow keys to examine the histogram.

Optional Collaborative Exercise

Count the money (bills and change) in your pocket or purse. Your instructor
will record the amounts. As a class, construct a histogram displaying the
data. Discuss how many intervals you think is appropriate. You may want to
experiment with the number of intervals. Discuss, also, the shape of the
histogram.

Record the data, in dollars (for example, 1.25 dollars).

Construct a histogram.

Glossary

Frequency
The number of times a value of the data occurs.

Relative Frequency
The ratio of the number of times a value of the data occurs in the set of
all outcomes to the number of all outcomes.

Frequency Polygons

Frequency polygons are a graphical device for understanding the shapes of
distributions. They serve the same purpose as histograms, but are especially
helpful in comparing sets of data. Frequency polygons are also a good
choice for displaying cumulative frequency distributions.

To create a frequency polygon, start just as for histograms, by choosing a
class interval. Then draw an X-axis representing the values of the scores in
your data. Mark the middle of each class interval with a tick mark, and label
it with the middle value represented by the class. Draw the Y-axis to
indicate the frequency of each class. Place a point in the middle of each
class interval at the height corresponding to its frequency. Finally, connect
the points. You should include one class interval below the lowest value in
your data and one above the highest value. The graph will then touch the X-
axis on both sides.

A frequency polygon for 642 psychology test scores is shown in [link]. The
first label on the X-axis is 35. This represents an interval extending from
29.5 to 39.5. Since the lowest test score is 46, this interval has a frequency
of 0. The point labeled 45 represents the interval from 39.5 to 49.5. There
are three scores in this interval. There are 150 scores in the interval that
surrounds 85.

You can easily discern the shape of the distribution from [link]. Most of the
scores are between 65 and 115. It is clear that the distribution is not
symmetric inasmuch as good scores (to the right) trail off more gradually
than poor scores (to the left). In the terminology of Chapter 3 (where we
will study shapes of distributions more systematically), the distribution is
skewed.

Frequency
oo BG
o oo oo o@ @&

a
o

26

35 45 55 65 75 85 95 105 115 125 135 145 155 165 175
Test Score

Frequency polygon for the psychology test scores.

A cumulative frequency polygon for the same test scores is shown in [Link].
The graph is the same as before except that the Y value for each point is the
number of students in the corresponding class interval plus all numbers in
lower intervals. For example, there are no scores in the interval labeled
"35," three in the interval "45,"and 10 in the interval "55."Therefore the Y
value corresponding to "55" is 13. Since 642 students took the test, the
cumulative frequency for the last interval is 642.

700
600
500
400

Cumulative Frequency

35 45 55 65 75 85 95 165 115 125 135 145 155 165
Test Score

Cumulative frequency polygon for the
psychology test scores.

Frequency polygons are useful for comparing distributions. This is achieved
by overlaying the frequency polygons drawn for different data sets. [link]
provides an example. The data come from a task in which the goal is to
move a computer mouse to a target on the screen as fast as possible. On 20
of the trials, the target was a small rectangle; on the other 20, the target was
a large rectangle. Time to reach the target was recorded on each trial. The
two distributions (one for each target) are plotted together in [link]. The
figure shows that although there is some overlap in times, it generally took
longer to move the mouse to the small target than to the large one.

large target

a

small target

Frequency

By

3568 456 5568 656 758 856 956 1656 1156

Time (msec)

Overlaid frequency polygons.

It is also possible to plot two cumulative frequency distributions in the same
graph. This is illustrated in [link] using the same data from the mouse task.
The difference in distributions for the two targets is again evident.

Cumulative Frequency

large target

small target

356 456 558 656 758 856 956

Time (msec)

Overlaid cumulative frequency polygons.

1856

Stem and Leaf Displays

A stem and leaf display is a graphical method of displaying data. It is
particularly useful when your data are not too numerous. In this section, we
will explain how to construct and interpret this kind of graph.

As usual, an example will get us started. Consider [link]. It shows the
number of touchdown (TD) passes [footnote] thrown by each of the 31
teams in the National Football League in the 2000 season.

Touchdown Pass: In American football, a touchdown pass occurs when a
completed pass results in a touchdown. The pass may be to a player in the
end zone or to a player who subsequently runs into the end zone. A
touchdown is worth 6 points and allows for a chance at one (and by some

rules two) additional point(s).

37, 33, 33, 32, 29, 28, 28, 23, 22, 22, 22, 21,
21,21, 20, 20, 19, 19, 18, 18, 18, 18, 16, 15,
14, 14, 14, 12, 12, 9,6

Number of touchdown passes.

A stem and leaf display of the data is shown in the [link] below. The left
portion of the table contains the stems. They are the numbers 3, 2, 1, and 0,
arranged as a column to the left of the bars. Think of these numbers as 10's
digits. A stem of 3 (for example) can be used to represent the 10's digit in
any of the numbers from 30 to 39. The numbers to the right of the bar are
leaves, and they represent the 1's digits. Every leaf in the graph therefore
stands for the result of adding the leaf to 10 times its stem.

3|2337

2/001112223889

1|2244456888899
0|69
Stem and leaf display showing the number of passing touchdowns.

To make this clear, let us examine this [link] more closely. In the top row,
the four leaves to the right of stem 3 are 2, 3, 3, and 7. Combined with the
stem, these leaves represent the numbers 32, 33, 33, and 37, which are the
numbers of TD passes for the first four teams in the table. The next row has
a stem of 2 and 12 leaves. Together, they represent 12 data points, namely,
two occurrences of 20 TD passes, three occurrences of 21 TD passes, three
occurrences of 22 TD passes, one occurrence of 23 TD passes, two
occurrences of 28 TD passes, and one occurrence of 29 TD passes. We
leave it to you to figure out what the third row represents. The fourth row
has a stem of 0 and two leaves. It stands for the last two entries, namely 9
TD passes and 6 TD passes. (The latter two numbers may be thought of as
09 and 06.).

One purpose of a stem and leaf display is to clarify the shape of the
distribution. You can see many facts about TD passes more easily in [link]
than in the [link]. For example, by looking at the stems and the shape of the
plot, you can tell that most of the teams had between 10 and 29 passing
TDs, with a few having more and a few having less. The precise numbers of
TD passes can be determined by examining the leaves.

We can make our figure even more revealing by splitting each stem into
two parts. The [link] below shows how to do this. The top row is reserved
for numbers from 35 to 39 and holds only the 37 TD passes made by the
first team in the [link]. The second row is reserved for the numbers from 30
to 34 and holds the 32, 33, and 33 TD passes made by the next three teams
in the table. You can see for yourself what the other rows represent.

3|7
3|233
2|889
2|001112223
1|56888899
122444
0|69
Stem and leaf display with the stems split in two.

The [link] with stem and leaf split in two is more revealing than the simpler
[link] before because the simpler table lumps too many values into a single

row. Whether you should split stems in a display depends on the exact form
of your data. If rows get too long with single stems, you might try splitting

them into two or more parts.

There is a variation of stem and leaf displays that is useful for comparing
distributions. The two distributions are placed back to back along a
common column of stems. The result is a back to back stem and leaf
graph. The [link] below shows such a graph. It compares the numbers of
TD passes in the 1998 and 2000 seasons. The stems are in the middle, the
leaves to the left are for the 1998 data, and the leaves to the right are for the
2000 data. For example, the second-to-last row shows that in 1998 there
were teams with 11, 12, and 13 TD passes, and in 2000 there were two
teams with 12 and three teams with 14 TD passes.

1998 2000

11 4
3 7

SoZ 3 233

8865 2 889
44331110 Z 001112223
987776665 1 56888899
321 1 22444
7 0 69

Back to back stem and leaf display. The left side shows the 1998 TD data
and the right side shows the 2000 TD data.

This [link] helps us see that the two seasons were similar, but that only in
1998 did any teams throw more than 40 TD passes.

There are two things about the football data that make them easy to graph
with stems and leaves. First, the data are limited to whole numbers that can
be represented with a one-digit stem and a one-digit leaf. Second, all the
numbers are positive. If the data include numbers with three or more digits,
or contain decimals, they can be rounded to two-digit accuracy. Negative
values are also easily handled. Let us look at another example.

[link] shows data from a study on aggressive thinking. Each value is the
mean difference over a series of trials between the time it took an
experimental subject to name aggressive words (like "punch") under two
conditions. In one condition the words were preceded by a non-weapon
word like "rabbit" or "bug." In the second condition, the same words were
preceded by a weapon word such as "gun" or "knife." The issue addressed

by the experiment was whether a preceding weapon word would speed up
(or prime) pronunciation of the aggressive word, compared to a non-
weapon priming word. A positive difference implies greater priming of the
aggressive word by the weapon word. Negative differences imply that the

priming by the weapon word was less than for a neutral word.
43.2, 42.9, 35.6, 35.1, 25.6, 25.4, 23.6, 205,

19.9, 14.4, 12.7, 11.3, 10.2, 10, 9.1, 7.5, 5.4,

47, 3.8, 2.1, 1.2, -0.2, -6.3, -6.7, -8.8, -10.4,

-10.5, -14.9, -14.9, -15, -18.5, -27.4

The effects of priming
(thousandths of a second).

You see that the numbers range from 43.2 to -27.4. The first value indicates
that one subject was 43.2 milliseconds faster pronouncing aggressive words
when they were preceded by weapon words than when preceded by neutral
words. The value -27.4 indicates that another subject was 27.4 milliseconds
slower pronouncing aggressive words when they were preceded by weapon
words.

The data are displayed with stems and leaves in the [link]. Since stem and
leaf displays can only portray two whole digits (one for the stem and one
for the leaf) the numbers are first rounded. Thus, the value 43.2 is rounded
to 43 and represented with a stem of 4 and a leaf of 3. Similarly, 42.9 is
rounded to 43. To represent negative numbers, we simply use negative
stems. For example, the bottom row of the figure represents the number -27.
The second-to-last row represents the numbers -10, -10, -15, etc. Once
again, we have rounded the original values from [link].

4|33

3|56

2|00456
100134
0|1245589
-0|0679
-1]005559
-2|7
Stem and leaf display with negative numbers and rounding

Observe that the figure contains a row headed by "0" and another headed
by"-0". The stem of 0 is for numbers between 0 and 9 whereas the stem of
-0 is for numbers between 0 and -9. For example, the fifth row of the table
holds the numbers 1, 2, 4, 5, 5, 8, 9 and the sixth row holds 0, -6, -7, and -9.
Values that are exactly 0 before rounding should be split as evenly as
possible between the "0" and "-0" rows. In [link], none of the values are 0
before rounding. The "0" that appears in the "-0" row comes from the
original value of -0.2 in the figure.

Although stem and leaf displays are unwieldy for large datasets, they are
often useful for datasets with up to 200 observations. [link] portrays the
distribution of populations of 185 US cities in 1998. To be included, a city
had to have between 100,000 and 500,000 residents.

|

|

| 444444444444555555555555

| 2222222222222222222333333333
|00000000000000011111111111111111111111111

Stem and leaf display of
populations of US cities with
populations between 100,000

and 500,000.

Since a stem and leaf plot shows only two-place accuracy, we had to round
the numbers to the nearest 10,000. For example the largest number
(493,559) was rounded to 490,000 and then plotted with a stem of 4 and a
leaf of 9. The fourth highest number (463,201) was rounded to 460,000 and
plotted with a stem of 4 and a leaf of 6. Thus, the stems represent units of
100,000 and the leaves represent units of 10,000. Notice that each stem
value is split into five parts: 0-1, 2-3, 4-5, 6-7, and 8-9.

Whether your data can be suitably represented by a stem and leaf graph
depends on whether they can be rounded without loss of important
information. Also, their extreme values must fit into two successive digits,
as the data in [link] fit into the 10,000 and 100,000 places (for leaves and
stems, respectively). Deciding what kind of graph is best suited to
displaying your data thus requires good judgment. Statistics is not just
recipes!

Box Plots

Box plots or box-whisker plots give a good graphical image of the
concentration of the data. They also show how far from most of the data the
extreme values are. The box plot is constructed from five values: the
smallest value, the first quartile, the median, the third quartile, and the
largest value. The median, the first quartile, and the third quartile will be
discussed here, and then again in the section on measuring data in this
chapter. We use these values to compare how close other data values are to
them.

The median, a number, is a way of measuring the "center" of the data. You
can think of the median as the "middle value," although it does not actually
have to be one of the observed values. It is a number that separates ordered
data into halves. Half the values are the same number or smaller than the
median and half the values are the same number or larger. For example,
consider the following data:

111.567.248910688322101
Ordered from smallest to largest:
11224668 7.288.39101011.5

The median is between the 7th value, 6.8, and the 8th value 7.2. To find the
median, add the two values together and divide by 2.
Equation:

6847.2 _

7
2

The median is 7. Half of the values are smaller than 7 and half of the values
are larger than 7.

Quartiles are numbers that separate the data into quarters. Quartiles may or
may not be part of the data. To find the quartiles, first find the median or
second quartile. The first quartile is the middle value of the lower half of

the data and the third quartile is the middle value of the upper half of the
data. To get the idea, consider the same data set shown above:

112246687.28839101011.5

The median or second quartile is 7. The lower half of the data is 1, 1, 2, 2,
4, 6, 6.8. The middle value of the lower half is 2.

11224668

The number 2, which is part of the data, is the first quartile. One-fourth of
the values are the same or less than 2 and three-fourths of the values are
more than 2.

The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of
the upper half is 9.

7.288.39 1010 11.5

The number 9, which is part of the data, is the third quartile. Three-fourths
of the values are less than 9 and one-fourth of the values are more than 9.

To construct a box plot, use a horizontal number line and a rectangular box.
The smallest and largest data values label the endpoints of the axis. The first
quartile marks one end of the box and the third quartile marks the other end
of the box. The middle fifty percent of the data fall inside the box. The
"whiskers" extend from the ends of the box to the smallest and largest data
values. The box plot gives a good quick picture of the data.

Note: You may encounter box and whisker plots that have dots marking
outlier values. In those cases, the whiskers are not extending to the
minimum and maximum values.

Consider the following data:

112246687.28839101011.5

The first quartile is 2, the median is 7, and the third quartile is 9. The
smallest value is 1 and the largest value is 11.5. The box plot is constructed
as follows (see calculator instructions in the back of this book or on the TI
web site):

1 2 K - 5 6 7 8 9 10 11 11.5

The two whiskers extend from the first quartile to the smallest value and
from the third quartile to the largest value. The median is shown with a
dashed line.

Example:

The following data are the heights of 40 students in a statistics class.

59 60 61 62 62 63 63 64 64 64 65 65 65 65 65 65 65 65 65 66 66 67 67 68
08 09! 70 70°70 70 70 71 71-72 72 73 7474 Ja 77

Construct a box plot:

Using the TI-83, 83+, 84, 84+ Calculator

e Enter data into the list editor (Press STAT 1:EDIT). If you need to
clear the list, arrow up to the name L1, press CLEAR, arrow down.

e Put the data values in list L1.

e Press STAT and arrow to CALC. Press 1:1-VarStats. Enter L1.

e Press ENTER

e Use the down and up arrow keys to scroll.

e Smallest value = 59
e Largest value = 77
e Q1: First quartile = 64.5

e Q2: Second quartile or median= 66
¢ Q3: Third quartile = 70

Using the TI-83, 83+, 84, 84+ to Construct the Box Plot
Go to 14:Appendix for Notes for the TI-83, 83+, 84, 84+ Calculator. To
create the box plot:

e Press Y=. If there are any equations, press CLEAR to clear them.

e Press 2nd Y=.

e Press 4:Plotsoff. Press ENTER

e Press 2nd Y=

e Press 1:Ploti. Press ENTER.

e Arrow down and then use the right arrow key to go to the 5th picture
which is the box plot. Press ENTER.

e Arrow down to Xlist: Press 2nd 1 for L1

e Arrow down to Freq: Press ALPHA. Press 1.

e Press ZOOM. Press 9:ZoomStat.

e Press TRACE and use the arrow keys to examine the box plot.

59 645 66 70 17

e aEach quarter has 25% of the data.

e bThe spreads of the four quarters are 64.5 - 59 = 5.5 (first quarter), 66
- 64.5 = 1.5 (second quarter), 70 - 66 = 4 (3rd quarter), and 77 - 70 =
7 (fourth quarter). So, the second quarter has the smallest spread and
the fourth quarter has the largest spread.

e cinterquartile Range: IQR = Q3 — Q1 = 70 — 64.5 = 5.5.

e dThe interval 59 through 65 has more than 25% of the data so it has
more data in it than the interval 66 through 70 which has 25% of the
data.

e eThe middle 50% (middle half) of the data has a range of 5.5 inches.

For some sets of data, some of the largest value, smallest value, first
quartile, median, and third quartile may be the same. For instance, you
might have a data set in which the median and the third quartile are the
same. In this case, the diagram would not have a dotted line inside the box
displaying the median. The right side of the box would display both the
third quartile and the median. For example, if the smallest value and the
first quartile were both 1, the median and the third quartile were both 5,
and the largest value was 7, the box plot would look as follows:

Example:

Test scores for a college statistics class held during the day are:

99 56 78 55.5 32 90 80 81 56 59 45 77 84.5 84 70 72 68 32 79 90

Test scores for a college statistics class held during the evening are:

98 78 68 83 81 89 88 76 65 45 98 90 80 84.5 85 79 78 98 90 79 81 25.5
Exercise:

Problem:

e What are the smallest and largest data values for each data set?

e¢ What is the median, the first quartile, and the third quartile for
each data set?

e Create a boxplot for each set of data.

e¢ Which boxplot has the widest spread for the middle 50% of the
data (the data between the first and third quartiles)? What does
this mean for that set of data in comparison to the other set of
data?

e For each data set, what percent of the data is between the
smallest value and the first quartile? (Answer: 25%) the first
quartile and the median? (Answer: 25%) the median and the third
quartile? the third quartile and the largest value? What percent of
the data is between the first quartile and the largest value?
(Answer: 75%)

Solution:
First Data Set

e Xmin = 32
«Oil 56

e M = 74.5
* (O3— 82:5
e Xmax = 99

Second Data Set

e Xmin = 25.5
Or 71s
M=8l

Q3 = 89
Xmax = 98

—L_i FF —

20 30 40 S50 60 70 80 90 100

The first data set (the top box plot) has the widest spread for the middle
50% of the data. IQR = Q3 — Q1 is 82.5 — 56 = 26.5 for the first data
set and 89 — 78 = 11 for the second data set. So, the first set of data has
its middle 50% of scores more spread out.

25% of the data is between M and Q3 and 25% is between Q3 and Xmax.

Glossary

Median
A number that separates ordered data into halves. Half the values are
the same number or smaller than the median and half the values are the
same number or larger than the median. The median may or may not
be part of the data.

Quartiles
The numbers that separate the data into quarters. Quartiles may or may
not be part of the data. The second quartile is the median of the data.

Measures of the Location of the Data

Descriptive Statistics: Measuring the Location of Data explains percentiles and
quartiles and is part of the collection col10555 written by Barbara Illowsky and
Susan Dean. Roberta Bloom contributed the section "Interpreting Percentiles,
Quartile and the Median."

The common measures of location are quartiles and percentiles (%iles). Quartiles
are special percentiles. The first quartile, Q ; is the same as the 25th percentile
(25th %ile) and the third quartile, @ 3, is the same as the 75th percentile (75th
%ile). The median, M, is called both the second quartile and the 50th percentile
(50th %ile).

Note: Quartiles are given special attention in the Box Plots module in this chapter.

To calculate quartiles and percentiles, the data must be ordered from smallest to
largest. Recall that quartiles divide ordered data into quarters. Percentiles divide
ordered data into hundredths. To score in the 90th percentile of an exam does not
mean, necessarily, that you received 90% on a test. It means that 90% of test
scores are the same or less than your score and 10% of the test scores are the same
or greater than your test score.

Percentiles are useful for comparing values. For this reason, universities and
colleges use percentiles extensively.

Percentiles are mostly used with very large populations. Therefore, if you were to
say that 90% of the test scores are less (and not the same or less) than your score,
it would be acceptable because removing one particular data value is not
significant.

The interquartile range is a number that indicates the spread of the middle half
or the middle 50% of the data. It is the difference between the third quartile (Q 3)
and the first quartile (Q 1).

Equation:

IQR = Q3— Qi

The IQR can help to determine potential outliers. A value is suspected to be a
potential outlier if it is less than (1.5)(IQR) below the first quartile or more
than (1.5)(IQR) above the third quartile. Potential outliers always need further
investigation.

Example:
Exercise:

Problem:

For the following 13 real estate prices, calculate the IQR and determine if
any prices are outliers. Prices are in dollars. (Source: San Jose Mercury
News)

389,950 230,500 158,000 479,000 639,000 114,950 5,500,000 387,000
659,000 529,000 575,000 488,800 1,095,000

Solution:

Order the data from smallest to largest.

114,950 158,000 230,500 387,000 389,950 479,000 488,800 529,000
975,000 639,000 659,000 1,095,000 5,500,000

M = 488,800

Qy = 20800387000 — 308750

Qz = $82000+859000 — 649000

IQR = 649000 — 308750 = 340250

(1.5)(IQR) = (1.5)(340250) = 510375

Qi — (1.5)(IQR) = 308750 — 510375 = —201625
Q3 + (1.5)(IQR) = 649000 + 510375 = 1159375

No house price is less than -201625. However, 5,500,000 is more than
1,159,375. Therefore, 5,500,000 is a potential outlier.

Example:
Exercise:

Problem:
For the two data sets in the test scores example, find the following:

e aThe interquartile range. Compare the two interquartile ranges.

e bAny outliers in either set.

¢ cThe 30th percentile and the 80th percentile for each set. How much
data falls below the 30th percentile? Above the 80th percentile?

Solution:

For the IQRs, see the answer to the test scores example. The first data set has
the larger IQR, so the scores between Q3 and Q1 (middle 50%) for the first
data set are more spread out and not clustered about the median.

First Data Set

¢ (2) - (IQR) = (2) - (26.5) = 39.75
e Xmax - Q3 = 99 - 82.5 = 1

(3) : (IQR) = 39.75 is larger than 16.5 and larger than 24, so the first set
has no outliers.

Second Data Set
(+) - (IQR) = (>) - (11) = 16.5
e Xmax — Q3 = 98 — 89 = 9
e QI — Xmin = 78 — 25.5 = 52.5

= : (IQR) = 16.5 is larger than 9 but smaller than 52.5, so for the
second set 45 and 25.5 are outliers.

To find the percentiles, create a frequency, relative frequency, and

Data Chapter). Get the percentiles from that chart.
First Data Set

¢ 30th %ile (between the 6th and 7th values) = pote) = iH

80th %ile (between the 16th and 17th values) = en) = 84.25
Second Data Set

¢ 30th %ile (7th value) = 78
° 80th %ile (18th value) = 90

30% of the data falls below the 30th %ile, and 20% falls above the 80th
%ile.

Example:

Finding Quartiles and Percentiles Using a Table

Fifty statistics students were asked how much sleep they get per school night
(rounded to the nearest hour). The results were (student data):

AMOUNT

OF

SLEEP

PER

SCHOOL CUMULATIVE
NIGHT RELATIVE RELATIVE
(HOURS) FREQUENCY FREQUENCY FREQUENCY

4 2 0.04 0.04

fs) fs) 0.10 0.14

AMOUNT

OF

SLEEP

PER

SCHOOL CUMULATIVE
NIGHT RELATIVE RELATIVE

(HOURS) FREQUENCY FREQUENCY FREQUENCY

6 7 0.14 0.28
ye 12 0.24 0.52
8 14 0.28 0.80
9 7. 0.14 0.94
10 3 0.06 1.00

Find the 28th percentile: Notice the 0.28 in the "cumulative relative frequency"
column. 28% of 50 data values = 14. There are 14 values less than the 28th %ile.
They include the two 4s, the five 5s, and the seven 6s. The 28th %ile is between
the last 6 and the first 7. The 28th %ile is 6.5.

Find the median: Look again at the "cumulative relative frequency " column and
find 0.52. The median is the 50th %ile or the second quartile. 50% of 50 = 25.
There are 25 values less than the median. They include the two 4s, the five 5s, the
seven 6s, and eleven of the 7s. The median or 50th %ile is between the 25th (7)
and 26th (7) values. The median is 7.

Find the third quartile: The third quartile is the same as the 75th percentile. You
can "eyeball" this answer. If you look at the "cumulative relative frequency"
column, you find 0.52 and 0.80. When you have all the 4s, 5s, 6s and 7s, you
have 52% of the data. When you include all the 8s, you have 80% of the data.
The 75th “%ile, then, must be an 8 . Another way to look at the problem is to
find 75% of 50 (= 37.5) and round up to 38. The third quartile, Q 3, is the 38th
value which is an 8. You can check this answer by counting the values. (There are
37 values below the third quartile and 12 values above.)

Example:

Exercise:

Problem: Using the table:

1. Find the 80th percentile.

2. Find the 90th percentile.

3. Find the first quartile.

4. What is another name for the first quartile?

Solution:

pe = BE
5 = 8.
Look where cum. rel. freg. = 0.80. 80% of the data is 8 or less. 80th
%ile is between the last 8 and first 9.
Das
Sie)
4. First Quartile = 25th %ile

Collaborative Classroom Exercise: Your instructor or a member of the class will
ask everyone in class how many sweaters they own. Answer the following
questions.

1. How many students were surveyed?

2. What kind of sampling did you do?

3. Construct a table of the data.

4. Construct 2 different histograms. For each, starting value = ending
value=

5. Use the table to find the median, first quartile, and third quartile.

6. Construct a box plot.

7. Use the table to find the following:

o The 10th percentile
o The 70th percentile
o The percent of students who own less than 4 sweaters

Interpreting Percentiles, Quartiles, and Median

A percentile indicates the relative standing of a data value when data are sorted
into numerical order, from smallest to largest. p% of data values are less than or
equal to the pth percentile. For example, 15% of data values are less than or equal
to the 15th percentile.

¢ Low percentiles always correspond to lower data values.
e High percentiles always correspond to higher data values.

A percentile may or may not correspond to a value judgment about whether it is
"good" or "bad". The interpretation of whether a certain percentile is good or bad
depends on the context of the situation to which the data applies. In some
situations, a low percentile would be considered "good'; in other contexts a high
percentile might be considered "good". In many situations, there is no value
judgment that applies.

Understanding how to properly interpret percentiles is important not only when
describing data, but is also important in later chapters of this textbook when
calculating probabilities.

Guideline:

When writing the interpretation of a percentile in the context of the given data, the
sentence should contain the following information:

e information about the context of the situation being considered,

e the data value (value of the variable) that represents the percentile,

e the percent of individuals or items with data values below the percentile.

e Additionally, you may also choose to state the percent of individuals or items
with data values above the percentile.

Example:
On a timed math test, the first quartile for times for finishing the exam was 35
minutes. Interpret the first quartile in the context of this situation.

e 25% of students finished the exam in 35 minutes or less.

e 75% of students finished the exam in 35 minutes or more.

e A low percentile could be considered good, as finishing more quickly on a
timed exam is desirable. (If you take too long, you might not be able to
finish.)

Example:
On a 20 question math test, the 70th percentile for number of correct answers was
16. Interpret the 70th percentile in the context of this situation.

¢ 70% of students answered 16 or fewer questions correctly.

e 30% of students answered 16 or more questions correctly.

e Note: A high percentile could be considered good, as answering more
questions correctly is desirable.

Example:

At a certain community college, it was found that the 30th percentile of credit
units that students are enrolled for is 7 units. Interpret the 30th percentile in the
context of this situation.

¢ 30% of students are enrolled in 7 or fewer credit units

e 70% of students are enrolled in 7 or more credit units

e In this example, there is no "good" or "bad" value judgment associated with
a higher or lower percentile. Students attend community college for varied
reasons and needs, and their course load varies according to their needs.

Do the following Practice Problems for Interpreting Percentiles
Exercise:

Problem:

e a For runners in arace, a low time means a faster run. The winners in a
race have the shortest running times. Is it more desirable to have a finish
time with a high or a low percentile when running a race?

¢ b The 20th percentile of run times in a particular race is 5.2 minutes.
Write a sentence interpreting the 20th percentile in the context of the
situation.

e cA bicyclist in the 90th percentile of a bicycle race between two towns
completed the race in 1 hour and 12 minutes. Is he among the fastest or
slowest cyclists in the race? Write a sentence interpreting the 90th
percentile in the context of the situation.

Solution:

e a Forrunners in a race it is more desirable to have a low percentile for
finish time. A low percentile means a short time, which is faster.

¢ bINTERPRETATION: 20% of runners finished the race in 5.2 minutes
or less. 80% of runners finished the race in 5.2 minutes or longer.

¢ cHe is among the slowest cyclists (90% of cyclists were faster than
him.) INTERPRETATION: 90% of cyclists had a finish time of 1 hour,
12 minutes or less.Only 10% of cyclists had a finish time of 1 hour, 12
minutes or longer

Exercise:

Problem:

e a For runners in a race, a higher speed means a faster run. Is it more
desirable to have a speed with a high or a low percentile when running a
race?

e bThe 40th percentile of speeds in a particular race is 7.5 miles per hour.
Write a sentence interpreting the 40th percentile in the context of the

situation.

Solution:

e aFor runners in a race it is more desirable to have a high percentile for
speed. A high percentile means a higher speed, which is faster.

¢ bINTERPRETATION: 40% of runners ran at speeds of 7.5 miles per
hour or less (slower). 60% of runners ran at speeds of 7.5 miles per hour
or more (faster).

Exercise:

Problem:

On an exam, would it be more desirable to earn a grade with a high or low
percentile? Explain.

Solution:
On an exam you would prefer a high percentile; higher percentiles
correspond to higher grades on the exam.

Exercise:
Problem:
Mina is waiting in line at the Department of Motor Vehicles (DMV). Her wait
time of 32 minutes is the 85th percentile of wait times. Is that good or bad?

Write a sentence interpreting the 85th percentile in the context of this
situation.

Solution:

When waiting in line at the DMV, the 85th percentile would be a long wait
time compared to the other people waiting. 85% of people had shorter wait
times than you did. In this context, you would prefer a wait time
corresponding to a lower percentile. INTERPRETATION: 85% of people at
the DMV waited 32 minutes or less. 15% of people at the DMV waited 32
minutes or longer.

Exercise:
Problem:
In a survey collecting data about the salaries earned by recent college

graduates, Li found that her salary was in the 78th percentile. Should Li be
pleased or upset by this result? Explain.

Solution:

Li should be pleased. Her salary is relatively high compared to other recent
college grads. 78% of recent college graduates earn less than Li does. 22% of
recent college graduates earn more than Li does.

Exercise:

Problem:

In a study collecting data about the repair costs of damage to automobiles in a
certain type of crash tests, a certain model of car had $1700 in damage and
was in the 90th percentile. Should the manufacturer and/or a consumer be
pleased or upset by this result? Explain. Write a sentence that interprets the
90th percentile in the context of this problem.

Solution:

The manufacturer and the consumer would be upset. This is a large repair
cost for the damages, compared to the other cars in the sample.
INTERPRETATION: 90% of the crash tested cars had damage repair costs of
$1700 or less; only 10% had damage repair costs of $1700 or more.

Exercise:

Problem:

e The University of California has two criteria used to set admission
standards for freshman to be admitted to a college in the UC system:

e a. Students' GPAs and scores on standardized tests (SATs and ACTs) are
entered into a formula that calculates an "admissions index" score. The
admissions index score is used to set eligibility standards intended to
meet the goal of admitting the top 12% of high school students in the
state. In this context, what percentile does the top 12% represent?

e b. Students whose GPAs are at or above the 96th percentile of all
students at their high school are eligible (called eligible in the local
context), even if they are not in the top 12% of all students in the state.
What percent of students from each high school are "eligible in the local
context"?

Solution:
¢ aThe top 12% of students are those who are at or above the 88th
percentile of admissions index scores.

¢ b The top 4% of students' GPAs are at or above the 96th percentile,
making the top 4% of students "eligible in the local context".

Exercise:

Problem:

Suppose that you are buying a house. You and your realtor have determined
that the most expensive house you can afford is the 34th percentile. The 34th
percentile of housing prices is $240,000 in the town you want to move to. In
this town, can you afford 34% of the houses or 66% of the houses?

Solution:

You can afford 34% of houses. 66% of the houses are too expensive for your
budget. INTERPRETATION: 34% of houses cost $240,000 or less. 66% of
houses cost $240,000 or more.

**With contributions from Roberta Bloom

Glossary

Interquartile Range (IRQ)
The distance between the third quartile (Q3) and the first quartile (Q1). IQR
= Q3 - Ql.

Outlier
An observation that does not fit the rest of the data.

Percentile
A number that divides ordered data into hundredths.

Example:
Let a data set contain 200 ordered observations starting with

{2.3,2.7,2.8,2.9,2.9,3.0...}. Then the first percentile is @7?*) — 9.75,

because 1% of the data is to the left of this point on the number line and 99% of

hae! eho 0n 2: ;
the data is on its right. The second percentile is ( = *) — 2.9. Percentiles may

or may not be part of the data. In this example, the first percentile is not in the
data, but the second percentile is. The median of the data is the second quartile
and the 50th percentile. The first and third quartiles are the 25th and the 75th
percentiles, respectively.

Quartiles
The numbers that separate the data into quarters. Quartiles may or may not be
part of the data. The second quartile is the median of the data.

Measures of Central Tendency
This chapter discusses measuring descriptive statistical information using the
center of the data

The "center" of a data set is also a way of describing location. The two most
widely used measures of the "center" of the data are the mean (average) and the
median. To calculate the mean weight of 50 people, add the 50 weights
together and divide by 50. To find the median weight of the 50 people, order
the data and find the number that splits the data into two equal parts (previously
discussed under box plots in this chapter). The median is generally a better
measure of the center when there are extreme values or outliers because it is not
affected by the precise numerical values of the outliers. The mean is the most
common measure of the center.

Note:The words "mean" and "average" are often used interchangeably. The
substitution of one word for the other is common practice. The technical term
is "arithmetic mean" and "average" is technically a center location. However,
in practice among non-statisticians, "average" is commonly accepted for
"arithmetic mean.”

The mean can also be calculated by multiplying each distinct value by its
frequency and then dividing the sum by the total number of data values. The
letter used to represent the sample mean is an x with a bar over it (pronounced
"e Dal ix,

The Greek letter js (pronounced "mew" ) represents the population mean. One
of the requirements for the sample mean to be a good estimate of the population
mean is for the sample taken to be truly random.

To see that both ways of calculating the mean are the same, consider the
sample:

11122344444
Equation:

1+14+1+24+24+34+44+4+4+4+4
OS SSS eS ee eS SS
11

2.7

Equation:

3x14+2x24+1x3+5x4
O—————————
11

2.0

In the second calculation for the sample mean, the frequencies are 3, 2, 1, and
De

You can quickly find the location of the median by using the expression —
The letter n is the total number of data values in the sample. If n is an odd
number, the median is the middle value of the ordered data (ordered smallest to
largest). If n is an even number, the median is equal to the two middle values
added together and divided by 2 after the data has been ordered. For example, if
the total number of data values is 97, then oa = —— = 49. The median is the
49th value in the ordered data. If the total number of data values is 100, then
igs = we = 50.5. The median occurs midway between the 50th and 51st
values. The location of the median and the value of the median are not the
same. The upper case letter (V is often used to represent the median. The next
example illustrates the location of the median and the value of the median.

Example:
Exercise:

Problem:

AIDS data indicating the number of months an AIDS patient lives after
taking a new antibody drug are as follows (smallest to largest):

3488 1011121314151516161717182122222424252626272729293132333
33434353740444447

Calculate the mean and the median.

Solution:

The calculation for the mean is:

= 23.6

ia [3+4+(8)(2)+10+11+12+13+414+(15)(2)+(16)(2)+...+35+37+40+(44) (2)+47|
<4 40

To find the median, M, first use the formula for the location. The location
is:

far —. sel __
oR

Starting at the smallest value, the median is located between the 20th and
21st values (the two 24s):

34881011121314151516161717182122222424
25262627272929313233333434353740444447

M= — =n

The median is 24.

Using the TI-83,83+,84, 84+ Calculators
Calculator Instructions are located in the menu item 14:Appendix (Notes for
the TI-83, 83+, 84, 84+ Calculators).

e Enter data into the list editor. Press STAT 1:EDIT

e Put the data values in list L1.

e Press STAT and arrow to CALC. Press 1:1-VarStats. Press 2nd 1 for L1
and ENTER.

e Press the down and up arrow keys to scroll.

ec = 23.6, M = 24

Example:
Exercise:

Problem:
Suppose that, in a small town of 50 people, one person earns $5,000,000

per year and the other 49 each earn $30,000. Which is the better measure
of the "center," the mean or the median?

Solution:

= es ae — 129400
M = 30000

(There are 49 people who earn $30,000 and one person who earns
$5,000,000.)

The median is a better measure of the "center" than the mean because 49
of the values are 30,000 and one is 5,000,000. The 5,000,000 is an outlier.
The 30,000 gives us a better sense of the middle of the data.

Another measure of the center is the mode. The mode is the most frequent
value. If a data set has two values that occur the same number of times, then the
set is bimodal.

Example:

Statistics exam scores for 20 students are as follows
Statistics exam scores for 20 students are as follows:
5055.59 59°63 63 72 72 72 72 72:76 7861 63:64 6484 90153
Exercise:

Problem:Find the mode.
Solution:

The most frequent score is 72, which occurs five times. Mode = 72.

Example:
Five real estate exam scores are 430, 430, 480, 480, 495. The data set is
bimodal because the scores 430 and 480 each occur twice.

When is the mode the best measure of the "center"? Consider a weight loss
program that advertises a mean weight loss of six pounds the first week of the
program. The mode might indicate that most people lose two pounds the first
week, making the program less appealing.

Note:The mode can be calculated for qualitative data as well as for
quantitative data.

Statistical software will easily calculate the mean, the median, and the mode.
Some graphing calculators can also make these calculations. In the real world,
people make these calculations using software.

The Law of Large Numbers and the Mean

The Law of Large Numbers says that if you take samples of larger and larger
size from any population, then the mean z of the sample is very likely to get
closer and closer to p. This is discussed in more detail in The Central Limit
Theorem.

Note:The formula for the mean is located in the Summary of Formulas section
course.

Sampling Distributions and Statistic of a Sampling Distribution

You can think of a sampling distribution as a relative frequency distribution
with a great many samples. (See Sampling and Data for a review of relative
frequency). Suppose thirty randomly selected students were asked the number
of movies they watched the previous week. The results are in the relative
frequency table shown below.

# of movies Relative Frequency

0

5/30

15/30

6/30

4/30

1/30

If you let the number of samples get very large (say, 300 million or more),
the relative frequency table becomes a relative frequency distribution.

A statistic is a number calculated from a sample. Statistic examples include the
mean, the median and the mode as well as others. The sample mean z is an
example of a statistic which estimates the population mean .

Glossary

Mean

A number that measures the central tendency. A common name for mean
is ‘average.’ The term 'mean' is a shortened form of ‘arithmetic mean.' By

definition, the mean for a sample (denoted by 2) is

Sum of all values in th 1 ,
ugbecot oslucsin theasmcle® and the mean for a population (denoted by

) isu = Sum of all values in the population
be /t = Number of values in the population *

|

Median

A number that separates ordered data into halves. Half the values are the
same number or smaller than the median and half the values are the same
number or larger than the median. The median may or may not be part of
the data.

Mode

The value that appears most frequently in a set of data.

Skewness and the Mean, Median, and Mode
Consider the following data set:
456667777778836910

This data set produces the histogram shown below. Each interval has width
one and each value is located in the middle of an interval.

4 5 6 7 8 9 10

The histogram displays a symmetrical distribution of data. A distribution is
symmetrical if a vertical line can be drawn at some point in the histogram
such that the shape to the left and the right of the vertical line are mirror
images of each other. The mean, the median, and the mode are each 7 for
these data. In a perfectly symmetrical distribution, the mean and the
median are the same. This example has one mode (unimodal) and the
mode is the same as the mean and median. In a symmetrical distribution
that has two modes (bimodal), the two modes would be different from the
mean and median.

The histogram for the data:
4566677778

is not symmetrical. The right-hand side seems "chopped off" compared to
the left side. The shape distribution is called skewed to the left because it is
pulled out to the left.

4 5 6 7 8

The mean is 6.3, the median is 6.5, and the mode is 7. Notice that the
mean is less than the median and they are both less than the mode. The
mean and the median both reflect the skewing but the mean more so.

The histogram for the data:
677778886910

is also not symmetrical. It is skewed to the right.

6 7 8 9 10

The mean is 7.7, the median is 7.5, and the mode is 7. Of the three statistics,
the mean is the largest, while the mode is the smallest. Again, the mean
reflects the skewing the most.

To summarize, generally if the distribution of data is skewed to the left, the
mean is less than the median, which is often less than the mode. If the
distribution of data is skewed to the right, the mode is often less than the
median, which is less than the mean.

Skewness and symmetry become important when we discuss probability
distributions in later chapters.

Measures of the Spread of the Data

Descriptive Statistics: Measuring the Spread of Data explains standard deviation as a measure of variation in data
and is part of the collection col10555 written by Barbara Illowsky and Susan Dean. Roberta Bloom made
contributions that helped to clarify the standard deviation and the variance.

An important characteristic of any set of data is the variation in the data. In some data sets, the data values are
concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean.
The most common measure of variation, or spread, is the standard deviation.

The standard deviation is a number that measures how far data values are from their mean.
The standard deviation

e provides a numerical measure of the overall amount of variation in a data set
e can be used to determine whether a particular data value is close to or far from the mean

The standard deviation provides a measure of the overall variation in a data set

The standard deviation is always positive or 0. The standard deviation is small when the data are all concentrated
close to the mean, exhibiting little variation or spread. The standard deviation is larger when the data values are
more spread out from the mean, exhibiting more variation.

Suppose that we are studying waiting times at the checkout line for customers at supermarket A and supermarket
B; the average wait time at both markets is 5 minutes. At market A, the standard deviation for the waiting time is 2
minutes; at market B the standard deviation for the waiting time is 4 minutes.

Because market B has a higher standard deviation, we know that there is more variation in the waiting times at
market B. Overall, wait times at market B are more spread out from the average; wait times at market A are more
concentrated near the average.

The standard deviation can be used to determine whether a data value is close to or far from the mean.
Suppose that Rosa and Binh both shop at Market A. Rosa waits for 7 minutes and Binh waits for 1 minute at the
checkout counter. At market A, the mean wait time is 5 minutes and the standard deviation is 2 minutes. The
standard deviation can be used to determine whether a data value is close to or far from the mean.

Rosa waits for 7 minutes:

e 7 is 2 minutes longer than the average of 5; 2 minutes is equal to one standard deviation.
e Rosa's wait time of 7 minutes is 2 minutes longer than the average of 5 minutes.
e Rosa's wait time of 7 minutes is one standard deviation above the average of 5 minutes.

Binh waits for 1 minute.

e 1is 4 minutes less than the average of 5; 4 minutes is equal to two standard deviations.

e Binh's wait time of 1 minute is 4 minutes less than the average of 5 minutes.

e Binh's wait time of 1 minute is two standard deviations below the average of 5 minutes.

e A data value that is two standard deviations from the average is just on the borderline for what many
statisticians would consider to be far from the average. Considering data to be far from the mean if it is more
than 2 standard deviations away is more of an approximate "rule of thumb" than a rigid rule. In general, the
shape of the distribution of the data affects how much of the data is further away than 2 standard deviations.
(We will learn more about this in later chapters.)

The number line may help you understand standard deviation. If we were to put 5 and 7 on a number line, 7 is to
the right of 5. We say, then, that 7 is one standard deviation to the right of 5 because
5 + (1)(2) =7.

If 1 were also part of the data set, then 1 is two standard deviations to the left of 5 because
5 + (-2)(2) =1.

—_———— I
0 123 45 67

e In general, a value = mean + (#ofSTDEV)(standard deviation)

e where #0fSTDEVs = the number of standard deviations

e 7 is one standard deviation more than the mean of 5 because: 7=5+(1)(2)
e 1is two standard deviations less than the mean of 5 because: 1=5+(—2)(2)

The equation value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a
population:

¢ sample: z = x + (##ofSTDEV)(s)
¢ Population: « = + (##ofSTDEV)(o)

The lower case letter s represents the sample standard deviation and the Greek letter o (sigma, lower case)
represents the population standard deviation.

The symbol a is the sample mean and the Greek symbol yu is the population mean.

Calculating the Standard Deviation

If x is a number, then the difference "x - mean" is called its deviation. In a data set, there are as many deviations
as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong
to a population, in symbols a deviation is x — yw . For sample data, in symbols a deviation is z— z .

The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are
data from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the
standard deviation depends on whether it is calculated from a population or a sample. The lower case letter s
represents the sample standard deviation and the Greek letter o (sigma, lower case) represents the population
standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate
of o.

To calculate the standard deviation, we need to calculate the variance first. The variance is an average of the
squares of the deviations (the x— x values for a sample, or the x — pu values for a population). The symbol o?
represents the population variance; the population standard deviation o is the square root of the population
variance. The symbol s? represents the sample variance; the sample standard deviation s is the square root of the
sample variance. You can think of the standard deviation as a special average of the deviations.

If the numbers come from a census of the entire population and not a sample, when we calculate the average of
the squared deviations to find the variance, we divide by N, the number of items in the population. If the data are
from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n-
1, one less than the number of items in the sample. You can see that in the formulas below.

Formulas for the Sample Standard Deviation
"Sie _ 2 2
» = [BEE ore = EE
e For the sample standard deviation, the denominator is n-1, that is the sample size MINUS 1.
Formulas for the Population Standard Deviation

2 2
ome J FEW) op g = J Et(e-w)
¢ For the population standard deviation, the denominator is N, the number of items in the population.

In these formulas, f represents the frequency with which a value appears. For example, if a value appears once, f
is 1. If a value appears three times in the data set or population, f is 3.

Sampling Variability of a Statistic

The statistic of a sampling distribution was discussed in Descriptive Statistics: Measuring the Center of the
Data. How much the statistic varies from one sample to another is known as the sampling variability of a
statistic. You typically measure the sampling variability of a statistic by its standard error. The standard error of
the mean is an example of a standard error. It is a special standard deviation and is known as the standard
deviation of the sampling distribution of the mean. You will cover the standard error of the mean in The Central

Limit Theorem (not now). The notation for the standard error of the mean is a where o is the standard

deviation of the population and n is the size of the sample.

Note: In practice, USE A CALCULATOR OR COMPUTER SOFTWARE TO CALCULATE THE
STANDARD DEVIATION. If you are using a TI-83,83+,84+ calculator, you need to select the appropriate
standard deviation o, or s; from the summary statistics. We will concentrate on using and interpreting the
information that the standard deviation gives us. However you should study the following step-by-step example to
help you understand how the standard deviation measures variation from the mean.

Example:

In a fifth grade class, the teacher was interested in the average age and the sample standard deviation of the ages
of her students. The following data are the ages fora SAMPLE of n = 20 fifth grade students. The ages are
rounded to the nearest half year:

8) 925 G5 IO WO) WO) WO) WO 3 WO! MOS Oss) Hi TL a AT aU aL TL. '55 TL 5) TL 5)

Equation:

9+95x2+10x4+10.5x4+11x6411.5~x3
—

= 10.525
20

The average age is 10.53 years, rounded to 2 places.
The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square
root of the variance. We will explain the parts of the table after calculating s.

Data Freq. Deviations Deviations?” (Freq.)(Deviations’)

x ud (x — x) (esa Ce =a)

9 1 9 — 10.525 = —1.525 (—1.525)? = 2.325625 1 x 2.325625 = 2.325625
9.5 2 9.5 — 10.525 = —1.025 (—1.025)” = 1.050625 2 x 1.050625 = 2.101250
10 4 10 — 10.525 = —0.525 (—0.525)” = 0.275625 4 x .275625 = 1.1025
10.5 4 10.5 — 10.525 = —0.025 (—0.025)” = 0.000625 4 x .000625 = .0025

11 6 11 — 10.525 = 0.475 (0.475)? = 0.225625 6 x .225625 = 1.35375

Data Freq. Deviations Deviations? (Freq.)(Deviations’)

11.5 3 11.5 — 10.525 = 0.975 (0.975)” = 0.950625 3 x .950625 = 2.851875

The sample variance, s”, is equal to the sum of the last column (9.7375) divided by the total number of data
values minus one (20 - 1):

s? = S85 — 0.5125

The sample standard deviation s is equal to the square root of the sample variance:

s = V0.5125 =. 0715891 Rounded to two decimal places, s = 0.72

Typically, you do the calculation for the standard deviation on your calculator or computer. The
intermediate results are not rounded. This is done for accuracy.

Exercise:

Problem: Verify the mean and standard deviation calculated above on your calculator or computer.

Solution:
Using the TI-83,83+,84+ Calculators

e Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the
name. Press CLEAR and arrow down.

e Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, 6, 3) into list

L2. Use the arrow keys to move around.

Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the

comma. Press ENTER.

e £=10.525

Use Sx because this is sample data (not a population): Sx=0.715891

For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation)
For a sample: z = x + (#ofSTDEVs)(s)

For a population: x = p + (#ofSTDEVs)( oc)

For this example, use xz = x + (#ofSTDEVs)(s) because the data is from a sample

Exercise:
Problem: Find the value that is 1 standard deviation above the mean. Find (x + 1s).

Solution:

(x + 1s) = 10.53 + (1)(0.72) = 11.25

Exercise:
Problem: Find the value that is two standard deviations below the mean. Find (x — 2s).

Solution:

(x — 2s) = 10.53 — (2)(0.72) = 9.09

Exercise:

Problem: Find the values that are 1.5 standard deviations from (below and above) the mean.

Solution:

° (x —1.5s) = 10.53 — (1.5)(0.72) = 9.45
° (« +1.5s) = 10.53 + (1.5)(0.72) = 11.61

Explanation of the standard deviation calculation shown in the table

The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than
is the data value 11. The deviations 0.97 and 0.47 indicate that. A positive deviation occurs when the data value is
greater than the mean. A negative deviation occurs when the data value is less than the mean; the deviation is
-1.525 for the data value 9. If you add the deviations, the sum is always zero. (For this example, there are n=20
deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you
make them positive numbers, and the sum will also be positive. The variance, then, is the average squared
deviation.

The variance is a squared measure and does not have the same units as the data. Taking the square root solves the
problem. The standard deviation measures the spread in the same units as the data.

Notice that instead of dividing by n=20, the calculation divided by n-1=20-1=19 because the data is a sample. For
the sample variance, we divide by the sample size minus one (n-1). Why not divide by n? The answer has to do
with the population variance. The sample variance is an estimate of the population variance. Based on the
theoretical mathematics that lies behind these calculations, dividing by (n-1) gives a better estimate of the
population variance.

Note: Your concentration should be on what the standard deviation tells us about the data. The standard deviation
is a number which measures how far the data are spread from the mean. Let a calculator or computer do the
arithmetic.

The standard deviation, s or a, is either zero or larger than zero. When the standard deviation is 0, there is no
spread; that is, the all the data values are equal to each other. The standard deviation is small when the data are all
concentrated close to the mean, and is larger when the data values show more variation from the mean. When the
standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make s
or o very large.

The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better "feel"
for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation
can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that
the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the
first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be
confusing, always graph your data.

Note:The formula for the standard deviation is at the end of the chapter.

Example:
Exercise:

Problem: Use the following data (first exam scores) from Susan Dean's spring pre-calculus class:

3342494953555561 6367686869697273 7478808388888890 929494949496100

e aCreate a chart containing the data, frequencies, relative frequencies, and cumulative relative
frequencies to three decimal places.
e bCalculate the following to one decimal place using a TI-83+ or TI-84 calculator:

iThe sample mean

iiThe sample standard deviation
iiiThe median

ivThe first quartile

vthe third quartile

vilQR

©). (Gb XO} ey 6)

e cConstruct a box plot and a histogram on the same set of axes. Make comments about the box plot, the
histogram, and the chart.

Solution:
ea

Data Frequency Relative Frequency Cumulative Relative Frequency
38) 1 0.032 0.032
42 1 0.032 0.064
49 2, 0.065 0.129
53 1 0.032 0.161
55 2: 0.065 0.226
61 il 0.032 0.258
63 1 0.032 0.29
67 1 0.032 0.322
68 2 0.065 0.387
69 2 0.065 0.452
72 il 0.032 0.484
73 1 0.032 0.516
74 1 0.032 0.548
78 1 0.032 0.580

80 1 0.032 0.612

Data Frequency Relative Frequency Cumulative Relative Frequency

83 1 0.032 0.644

88 3 0.097 0.741

90 1 0.032 O).7/7/3!

92 il 0.032 0.805

94 4 0.129 0.934

96 il 0.032 0.966

100 il 0.032 0.998 (Why isn't this value 1?)
eb

o iThe sample mean = 73.5

o iiThe sample standard deviation = 17.9

o iiiThe median = 73

© ivThe first quartile = 61

o vThe third quartile = 90

© vilIQR = 90 - 61 = 29

e cThe x-axis goes from 32.5 to 100.5; y-axis goes from -2.4 to 15 for the histogram; number of intervals
is 5 for the histogram so the width of an interval is (100.5 - 32.5) divided by 5 which is equal to 13.6.
Endpoints of the intervals: starting point is 32.5, 32.5+13.6 = 46.1, 46.1+13.6 = 59.7, 59.7+13.6 = 73.3,
73.3+13.6 = 86.9, 86.9+13.6 = 100.5 = the ending value; No data values fall on an interval boundary.

The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in
the lower 50% is greater (73 - 33 = 40) than the spread in the upper 50% (100 - 73 = 27). The histogram, box plot,
and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram
clearly shows this. The box plot shows us that the middle 50% of the exam scores (IQR = 29) are Ds, Cs, and Bs.
The box plot also shows us that the lower 25% of the exam scores are Ds and Fs.

Comparing Values from Different Data Sets
The standard deviation is useful when comparing data values that come from different data sets. If the data sets
have different means and standard deviations, it can be misleading to compare the data values directly.

e For each data value, calculate how many standard deviations the value is away from its mean.
e Use the formula: value = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs.
#ofSTDEVs _ value—mean

standard deviation
¢ Compare the results of this calculation.

#ofSTDEVs is often called a "z-score"; we can use the symbol z. In symbols, the formulas become:

= — &£-2z
Sample L=XH+ZS a ae
: _ _ op
Population L=p~+zo z=
Example:
Exercise:
Problem:

Two students, John and Ali, from different high schools, wanted to find out who had the highest G.P.A. when
compared to his school. Which student had the highest G.P.A. when compared to his school?

Student GPA School Mean GPA School Standard Deviation
John 2.85 3.0 0.7
Ali 77 80 10

Solution:

For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average,
for his school. Pay careful attention to signs when comparing and interpreting the answer.

#ofSTDEVs = value—mean [c= TU

standard deviation o
For John, z = ##ofSTDEVs = 243-3° — —0.21
For Ali, z = #ofSTDEVs = Ts = —0.3

John has the better G.P.A. when compared to his school because his G.P.A. is 0.21 standard deviations below
his school's mean while Ali's G.P.A. is 0.3 standard deviations below his school's mean.

John's z-score of —0.21 is higher than Ali's z-score of —0.3 . For GPA, higher values are better, so we
conclude that John has the better GPA when compared to his school.

The following lists give a few facts that provide a little more insight into what the standard deviation tells us about
the distribution of the data.
For ANY data set, no matter what the distribution of the data is:

e At least 75% of the data is within 2 standard deviations of the mean.

e At least 89% of the data is within 3 standard deviations of the mean.

e At least 95% of the data is within 4 1/2 standard deviations of the mean.
e This is known as Chebyshev's Rule.

For data having a distribution that is MOUND-SHAPED and SYMMETRIC:

e Approximately 68% of the data is within 1 standard deviation of the mean.

e Approximately 95% of the data is within 2 standard deviations of the mean.

¢ More than 99% of the data is within 3 standard deviations of the mean.

e This is known as the Empirical Rule.

e It is important to note that this rule only applies when the shape of the distribution of the data is mound-
shaped and symmetric. We will learn more about this when studying the "Normal" or "Gaussian" probability
distribution in later chapters.

**With contributions from Roberta Bloom

Glossary

Standard Deviation
A number that is equal to the square root of the variance and measures how far data values are from their
mean. Notation: s for sample standard deviation and o for population standard deviation.

Variance
Mean of the squared deviations from the mean. Square of the standard deviation. For a set of data, a deviation
can be represented as x — x where z is a value of the data and z is the sample mean. The sample variance is
equal to the sum of the squares of the deviations divided by the difference of the sample size and 1.

Summary of Formulas
A summary of useful formulas used in examining descriptive statistics
Commonly Used Symbols

e The symbol »’ means to add or to find the sum.

e n= the number of data values in a sample

e N =the number of people, things, etc. in the population
e «x =the sample mean

e s =the sample standard deviation

e y= the population mean

e o =the population standard deviation

e f = frequency

e x = numerical value

Commonly Used Expressions

e x f =A value multiplied by its respective frequency

e )\ a = The sum of the values

e Sa f =The sum of values multiplied by their respective frequencies

* x 2 or £- yp = Deviations from the mean (how far a value is
from the mean)

°* x «x or x p = Deviations squared

efx « orf x yp =The deviations squared and multiplied by
their frequencies

Mean Formulas:

° 2x 2 org Lie
de fax

oh ARO OF B= Sy

Standard Deviation Formulas:

Sx“£ 2£ Sif «2 2
es A or s =a
ax pb “Sf x2 p

ay ORO

Formulas Relating a Value, the Mean, and the Standard Deviation:

e value = mean + (#ofSTDEVs)(standard deviation), where #ofSTDEVs
= the number of standard deviations

e x = x£+ (HofSTDEVs)(s)

e x = p+ (#ofSTDEVs)(oc)

Practice 1: Center of the Data

This module provides students with opportunities to apply concepts related
to descriptive statistics. Students are asked to take a set of sample data and
calculate a series of statistical values for that data.

Student Learning Outcomes

e The student will calculate and interpret the center, spread, and location
of the data.
e The student will construct and interpret histograms an box plots.

Given

Sixty-five randomly selected car salespersons were asked the number of
cars they generally sell in one week. Fourteen people answered that they
generally sell three cars; nineteen generally sell four cars; twelve generally
sell five cars; nine generally sell six cars; eleven generally sell seven cars.

Complete the Table

Cumulative
Data Value Relative Relative
(# cars) Frequency Frequency Frequency

Discussion Questions

Exercise:

Problem: What does the frequency column sum to? Why?

Solution:
65

Exercise:

Problem: What does the relative frequency column sum to? Why?

Solution:

1
Exercise:

Problem:

What is the difference between relative frequency and frequency for
each data value?

Exercise:

Problem:

What is the difference between cumulative relative frequency and
relative frequency for each data value?

Enter the Data

Enter your data into your calculator or computer.

Construct a Histogram

Determine appropriate minimum and maximum x and y values and the
scaling. Sketch the histogram below. Label the horizontal and vertical axes
with words. Include numerical scaling.

Data Statistics

Calculate the following values:
Exercise:

Problem: Sample mean = Z =

Solution:

4.75

Exercise:

Problem: Sample standard deviation = s, =

Solution:

1.39

Exercise:

Problem: Sample size = n =

Solution:

65

Calculations

Use the table in section 2.11.3 to calculate the following values:
Exercise:

Problem: Median =

Solution:
4
Exercise:

Problem: Mode =

Solution:
4
Exercise:

Problem: First quartile =

Solution:
4
Exercise:

Problem: Second quartile = median = 50th percentile =

Solution:

4

Exercise:

Problem: Third quartile =

Solution:
6

Exercise:

Problem: Interquartile range (IQR) = en
Solution:
6—4=2

Exercise:

Problem: 10th percentile =

Solution:
3
Exercise:

Problem: 70th percentile =

Solution:

6

Exercise:

Problem: Find the value that is 3 standard deviations:
e aAbove the mean
e bBelow the mean

Solution:

e a8.93

e b0.58

Box Plot

Construct a box plot below. Use a ruler to measure and scale accurately.

Interpretation

Looking at your box plot, does it appear that the data are concentrated
together, spread out evenly, or concentrated in some areas, but not in
others? How can you tell?

Practice 2: Spread of the Data
Practice exercise for Descriptive Statistics

Student Learning Outcomes

e The student will calculate measures of the center of the data.
e The student will calculate the spread of the data.

Given

The population parameters below describe the full-time equivalent number of
students (FTES) each year at Lake Tahoe Community College from 1976-77
through 2004-2005. (Source: Graphically Speaking by Bill King, LTCC
Institutional Research, December 2005).

Use these values to answer the following questions:

e «= 1000 FTES

¢ Median = 1014 FTES

e g =474FTES

e First quartile = 528.5 FTES

e Third quartile = 1447.5 FTES
e n= 29 years

Calculate the Values

Exercise:

Problem:

A sample of 11 years is taken. About how many are expected to have a
FTES of 1014 or above? Explain how you determined your answer.

Solution:

6

Exercise:

Problem: 75% of all years have a FTES:
e aAt or below:
e bAt or above:

Solution:

e al447.5
¢ b528.5

Exercise:

Problem: The population standard deviation =

Solution:

474 FTES
Exercise:

Problem:

What percent of the FTES were from 528.5 to 1447.5? How do you
know?

Solution:
50%
Exercise:
Problem: What is the IQR? What does the IQR represent?
Solution:

319

Exercise:

Problem:
How many standard deviations away from the mean is the median?
Solution:

0.03

Additional Information: The population FTES for 2005-2006 through 2010-
2011 was given in an updated report. (Source:
http://www.ltcc.edu/data/ResourcePDF/LTCC_FactBook_2010-11.pdf). The
data are reported here.

2005- 2006- 2007- 2008- 2009- 2010-

xear 06 07 08 09 10 14
Total 1585 1690 1735 1935 2021 1890
FTES

Exercise:
Problem:

Calculate the mean, median, standard deviation, first quartile, the third
quartile and the IQR. Round to one decimal place.

Solution:

mean = 1809.3

median = 1812.5

standard deviation = 151.2
First quartile = 1690

Third quartile = 1935
TQR = 245

Exercise:
Problem:
Construct a boxplot for the FTES for 2005-2006 through 2010-2011 and
a boxplot for the FTES for 1976-1977 through 2004-2005.
Exercise:
Problem:
Compare the IQR for the FTES for 1976-77 through 2004-2005 with the

IQR for the FTES for 2005-2006 through 2010-2011. Why do you
suppose the IQRs are so different?

Solution:

Hint: Think about the number of years covered by each time period and
what happened to higher education during those periods.

Homework

Descriptive Statistics: Homework is part of the collection col10555 written
by Barbara Illowsky and Susan Dean and provides homework questions
related to lessons about descriptive statistics.

Exercise:

Problem:

Twenty-five randomly selected students were asked the number of
movies they watched the previous week. The results are as follows:

Cumulative

# of Relative Relative
movies Frequency Frequency Frequency
0 s)

1 9

2 6

3 4

4 1

aFind the sample mean &

bFind the sample standard deviation, s
cConstruct a histogram of the data.
dComplete the columns of the chart.
eFind the first quartile.

fFind the median.

gFind the third quartile.

hConstruct a box plot of the data.

e iWhat percent of the students saw fewer than three movies?
e jFind the 40th percentile.

e kFind the 90th percentile.

e IConstruct a line graph of the data.

e mConstruct a stem plot of the data.

Solution:

al.48
b1.12
el
fl

e 180%
e jl
e k3

Exercise:

Problem:

The median age for U.S. blacks currently is 30.9 years; for U.S. whites
it is 42.3 years. ((Source:
http://www.usatoday.com/news/nation/story/2012-05-17/minority-
births-census/55029100/1))

e aBased upon this information, give two reasons why the black
median age could be lower than the white median age.

¢ bDoes the lower median age for blacks necessarily mean that
blacks die younger than whites? Why or why not?

¢ cHow might it be possible for blacks and whites to die at
approximately the same age, but for the median age for whites to
be higher?

Exercise:
Problem:
Forty randomly selected students were asked the number of pairs of

sneakers they owned. Let X = the number of pairs of sneakers owned.
The results are as follows:

Relative Cumulative Relative

xX Frequency Frequency Frequency

1 2

2 5

3 8

4 12

is) 12

7 1

e aFind the sample mean z

e bFind the sample standard deviation, s
e cConstruct a histogram of the data.

e dComplete the columns of the chart.

e eFind the first quartile.

fFind the median.

gFind the third quartile.

hConstruct a box plot of the data.

iWhat percent of the students owned at least five pairs?
e jFind the 40th percentile.

e kFind the 90th percentile.

e IConstruct a line graph of the data

e mConstruct a stem plot of the data

Solution:

e a3.78
e b1.29
e e3
e £4
e go

e 132.5%
e j4
e k5
Exercise:
Problem:
600 adult Americans were asked by telephone poll, What do you think
constitutes a middle-class income? The results are below. Also, include

left endpoint, but not the right endpoint. (Source: Time magazine;
survey by Yankelovich Partners, Inc.)

Note:"Not sure" answers were omitted from the results.

Salary ($) Relative Frequency
< 20,000 0.02
20,000 - 25,000 0.09
25,000 - 30,000 0.19
30,000 - 40,000 0.26
40,000 - 50,000 0.18
50,000 - 75,000 0.17
75,000 - 99,999 0.02
100,000+ 0.01

e aWhat percent of the survey answered "not sure" ?
e bWhat percent think that middle-class is from $25,000 - $50,000
i)

e cConstruct a histogram of the data

1. iShould all bars have the same width, based on the data?
Why or why not?

2. litHow should the <20,000 and the 100,000+ intervals be
handled? Why?

e dFind the 40th and 80th percentiles
e eConstruct a bar graph of the data

Exercise:

Problem:

Following are the published weights (in pounds) of all of the team
members of the San Francisco 49ers from a previous year (Source: San
Jose Mercury News)

177 205 210 210 232 205 185 185 178 210 206 212 184 174 185 242
188 212 215 247 241 223 220 260 245 259 278 270 280 295 275 285
290 272 273 280 285 286 200 215 185 230 250 241 190 260 250 302
265 290 276 228 265

aOrganize the data from smallest to largest value.

bFind the median.

cFind the first quartile.

dFind the third quartile.

eConstruct a box plot of the data.

fThe middle 50% of the weights are from to

gIf our population were all professional football players, would
the above data be a sample of weights or the population of
weights? Why?

hIf our population were the San Francisco 49ers, would the above
data be a sample of weights or the population of weights? Why?
iAssume the population was the San Francisco 49ers. Find:

ithe population mean, j.

iithe population standard deviation, o.

iiithe weight that is 2 standard deviations below the mean.
ivWhen Steve Young, quarterback, played football, he
weighed 205 pounds. How many standard deviations above
or below the mean was he?

QO -O -9:-0

jThat same year, the mean weight for the Dallas Cowboys was
240.08 pounds with a standard deviation of 44.38 pounds. Emmit
Smith weighed in at 209 pounds. With respect to his team, who

was lighter, Smith or Young? How did you determine your
answer?

Solution:

e b241
e ¢205.5
e d272.5
ee

174 205.5 241 272.5 302

e £205.5, 272.5
e gsample

e hpopulation
ej

1236.34

1137.50

i1161.34

iv0.84 std. dev. below the mean

o Oo 0 90

e jyYoung

Exercise:

Problem:

An elementary school class ran 1 mile with a mean of 11 minutes and a
standard deviation of 3 minutes. Rachel, a student in the class, ran 1
mile in 8 minutes. A junior high school class ran 1 mile with a mean of
9 minutes and a standard deviation of 2 minutes. Kenji, a student in the
class, ran 1 mile in 8.5 minutes. A high school class ran 1 mile with a
mean of 7 minutes and a standard deviation of 4 minutes. Nedda, a
student in the class, ran 1 mile in 8 minutes.

e aWhy is Kenji considered a better runner than Nedda, even
though Nedda ran faster than he?

e bWho is the fastest runner with respect to his or her class?
Explain why.

Exercise:

Problem:

In a survey of 20 year olds in China, Germany and America, people
were asked the number of foreign countries they had visited in their
lifetime. The following box plots display the results.

China

ee a ee

0 1 2 3 4 3 6 7 8 9 10 11

e aln complete sentences, describe what the shape of each box plot
implies about the distribution of the data collected.

e bExplain how it is possible that more Americans than Germans
surveyed have been to over eight foreign countries.

e cCompare the three box plots. What do they imply about the
foreign travel of twenty year old residents of the three countries
when compared to each other?

Exercise:

Problem:

One hundred teachers attended a seminar on mathematical problem
solving. The attitudes of a representative sample of 12 of the teachers
were measured before and after the seminar. A positive number for
change in attitude indicates that a teacher's attitude toward math
became more positive. The twelve change scores are as follows:

38-1205-31-165-2

e aWhat is the mean change score?

¢ bWhat is the standard deviation for this population?

e cWhat is the median change score?

e dFind the change score that is 2.2 standard deviations below the
mean.

Exercise:

Problem:

Three students were applying to the same graduate school. They came
from schools with different grading systems. Which student had the
best G.P.A. when compared to his school? Explain how you
determined your answer.

School Ave. School Standard

Student G.P.A. G.P.A. Deviation
Thuy 2) 3.2 0.8
Vichet 87 75 20
Kamala 8.6 8 0.4
Solution:
Kamala
Exercise:

Problem: Given the following box plot:

0 2 10 12 13

e aWhich quarter has the smallest spread of data? What is that
spread?

e bWhich quarter has the largest spread of data? What is that
spread?

e cFind the Inter Quartile Range (IQR).

e dAre there more data in the interval 5 - 10 or in the interval 10 -
13? How do you know this?

e eWhich interval has the fewest data in it? How do you know this?

I 0-2
II2-4
TIT10-12
IV12-13

o Oo 0 0

Exercise:

Problem: Given the following box plot:

0 20 100 150

e aThink of an example (in words) where the data might fit into the
above box plot. In 2-5 sentences, write down the example.

e bWhat does it mean to have the first and second quartiles so close
together, while the second to fourth quartiles are far apart?

Exercise:
Problem:

Santa Clara County, CA, has approximately 27,873 Japanese-
Americans. Their ages are as follows. (Source: West magazine)

Age Group Percent of Community
0-17 18.9

18-24 8.0

25-34 22.8

35-44 15.0

45-54 13.1

Age Group Percent of Community
55-64 11.9

65+ 10.3

e aConstruct a histogram of the Japanese-American community in
Santa Clara County, CA. The bars will not be the same width for
this example. Why not?

e¢ bWhat percent of the community is under age 35?

e cWhich box plot most resembles the information above?

0 24 34 53 =100
ii.
0 18 34 45 =100
iii.
0 24 25 54 =100
Exercise:
Problem:

Suppose that three book publishers were interested in the number of
fiction paperbacks adult consumers purchase per month. Each
publisher conducted a survey. In the survey, each asked adult
consumers the number of fiction paperbacks they had purchased the
previous month. The results are below.

# of books Freq. Rel. Freq.

0 10
1 12
2 16
3 12
4 8
5 6
6 yi
8 2
Publisher A

# of books Freq. Rel. Freq.
0 18
1 24
2 24
3 22

# of books Freq. Rel. Freq.

5 10
7 5
9 1
Publisher B
# of books Freq. Rel. Freq.
0-1 20
2-3 35
4-5 12
6-7 2
8-9 1
Publisher C

e aFind the relative frequencies for each survey. Write them in the
charts.

e bUsing either a graphing calculator, computer, or by hand, use the
frequency column to construct a histogram for each publisher's
survey. For Publishers A and B, make bar widths of 1. For
Publisher C, make bar widths of 2.

e cIn complete sentences, give two reasons why the graphs for
Publishers A and B are not identical.

e dWould you have expected the graph for Publisher C to look like
the other two graphs? Why or why not?

e eMake new histograms for Publisher A and Publisher B. This
time, make bar widths of 2.

e fNow, compare the graph for Publisher C to the new graphs for
Publishers A and B. Are the graphs more similar or more
different? Explain your answer.

Exercise:

Problem:

Often, cruise ships conduct all on-board transactions, with the
exception of gambling, on a cashless basis. At the end of the cruise,
guests pay one bill that covers all on-board transactions. Suppose that
60 single travelers and 70 couples were surveyed as to their on-board
bills for a seven-day cruise from Los Angeles to the Mexican Riviera.
Below is a summary of the bills for each group.

Amount($)
51-100
101-150
151-200
201-250
251-300

301-350

Frequency Rel. Frequency
5

10

15

15

10

Singles

Amount($) Frequency Rel. Frequency
100-150 fs)
201-250 is)
251-300 fs)
301-350 is)
351-400 10
401-450 10
451-500 10
501-550 10
551-600 fs)
601-650 fs)
Couples

e aFill in the relative frequency for each group.

e bConstruct a histogram for the Singles group. Scale the x-axis by
$50. widths. Use relative frequency on the y-axis.

e cConstruct a histogram for the Couples group. Scale the x-axis by
$50. Use relative frequency on the y-axis.

e dCompare the two graphs:

© iList two similarities between the graphs.
o jiList two differences between the graphs.
o iiiOverall, are the graphs more similar or different?

e eConstruct a new graph for the Couples by hand. Since each
couple is paying for two individuals, instead of scaling the x-axis
by $50, scale it by $100. Use relative frequency on the y-axis.

e fCompare the graph for the Singles with the new graph for the
Couples:

© iList two similarities between the graphs.
© iiOverall, are the graphs more similar or different?

e iBy scaling the Couples graph differently, how did it change the
way you compared it to the Singles?

e jBased on the graphs, do you think that individuals spend the
Same amount, more or less, as singles as they do person by person
in a couple? Explain why in one or two complete sentences.

Exercise:
Problem:
Refer to the following histograms and box plot. Determine which of

the following are true and which are false. Explain your solution to
each part in complete sentences.

1 3 6

aThe medians for all three graphs are the same.

bWe cannot determine if any of the means for the three graphs is
different.

cThe standard deviation for (b) is larger than the standard
deviation for (a).

dWe cannot determine if any of the third quartiles for the three
graphs is different.

Solution:
e alrue
e bTrue

e clrue
e dFalse

Exercise:

Problem: Refer to the following box plots.

Data 1

0 ys 4 i |
Data 2

0 2 7

e aln complete sentences, explain why each statement is false.

°o iData 1 has more data values above 2 than Data 2 has above
2,

o jiThe data sets cannot have the same mode.

o iiiFor Data 1, there are more data values below 4 than there
are above 4.

¢ bFor which group, Data 1 or Data 2, is the value of “7” more
likely to be an outlier? Explain why in complete sentences

Exercise:

Problem:

In a recent issue of the IEEE Spectrum, 84 engineering conferences
were announced. Four conferences lasted two days. Thirty-six lasted
three days. Eighteen lasted four days. Nineteen lasted five days. Four
lasted six days. One lasted seven days. One lasted eight days. One
lasted nine days. Let X = the length (in days) of an engineering
conference.

e aOrganize the data in a chart.

e bFind the median, the first quartile, and the third quartile.

e cFind the 65th percentile.

e dFind the 10th percentile.

e eConstruct a box plot of the data.

e fThe middle 50% of the conferences last from days to

days.

e gCalculate the sample mean of days of engineering conferences.

e hCalculate the sample standard deviation of days of engineering
conferences.

e iFind the mode.

e jlf you were planning an engineering conference, which would
you choose as the length of the conference: mean; median; or
mode? Explain why you made that choice.

e kGive two reasons why you think that 3 - 5 days seem to be
popular lengths of engineering conferences.

Solution:

e b4,3,5
e c4
e d3
ee

° £3,5
g3.94
h1.28
i3
jmode

Exercise:

Problem:

A survey of enrollment at 35 community colleges across the United
States yielded the following figures (source: Microsoft Bookshelf):

6414 1550 2109 9350 21828 4300 5944 5722 2825 2044 5481 5200
9853 2750 10012 6357 27000 9414 7681 3200 17500 9200 7380
18314 6557 13713 17768 7493 2771 2861 1263 7285 28165 5080
11622

e aOrganize the data into a chart with five intervals of equal width.
Label the two columns "Enrollment" and "Frequency."

¢ bConstruct a histogram of the data.

e clf you were to build a new community college, which piece of
information would be more valuable: the mode or the mean?

e dCalculate the sample mean.

e eCalculate the sample standard deviation.

e fA school with an enrollment of 8000 would be how many
standard deviations away from the mean?

Exercise:

Problem:

The median age of the U.S. population in 1980 was 30.0 years. In
1991, the median age was 33.1 years. (Source: Bureau of the Census)

e¢ aWhat does it mean for the median age to rise?

e bGive two reasons why the median age could rise.

e cFor the median age to rise, is the actual number of children less
in 1991 than it was in 1980? Why or why not?

Solution:

e cMaybe

Exercise:

Problem:

A survey was conducted of 130 purchasers of new BMW 3 series cars,
130 purchasers of new BMW 5 series cars, and 130 purchasers of new
BMW 7 series cars. In it, people were asked the age they were when
they purchased their car. The following box plots display the results.

BMW 3 series

25 30 35 40 45 50 55 60 65 70 75 80

e aln complete sentences, describe what the shape of each box plot
implies about the distribution of the data collected for that car

series.

e bWhich group is most likely to have an outlier? Explain how you
determined that.

e cCompare the three box plots. What do they imply about the age
of purchasing a BMW from the series when compared to each
other?

e dLook at the BMW 5 series. Which quarter has the smallest
spread of data? What is that spread?

e eLook at the BMW 5 series. Which quarter has the largest spread
of data? What is that spread?

e fLook at the BMW 5 series. Estimate the Inter Quartile Range
(IQR).

e gLook at the BMW 5 series. Are there more data in the interval
31-38 or in the interval 45-55? How do you know this?

e hLook at the BMW 5 series. Which interval has the fewest data in
it? How do you know this?

° 431-35
© 1138-41
° 11141-64

Exercise:
Problem:

The following box plot shows the U.S. population for 1990, the latest
available year. (Source: Bureau of the Census, 1990 Census)

0 17 33 50 #105

e aAre there fewer or more children (age 17 and under) than senior
citizens (age 65 and over)? How do you know?

e b12.6% are age 65 and over. Approximately what percent of the
population are of working age adults (above age 17 to age 65)?

Solution:

e amore children
e b62.4%

Exercise:

Problem:

Javier and Ercilia are supervisors at a shopping mall. Each was given
the task of estimating the mean distance that shoppers live from the
mall. They each randomly surveyed 100 shoppers. The samples
yielded the following information:

Javier Ercilla
x 6.0 miles 6.0 miles
S 4.0 miles 7.0 miles

e aHow can you determine which survey was correct ?
e bExplain what the difference in the results of the surveys implies

about the data.
clf the two histograms depict the distribution of values for each

supervisor, which one depicts Ercilia's sample? How do you
know?

e dif the two box plots depict the distribution of values for each
supervisor, which one depicts Ercilia’s sample? How do you

know?
i. ii.
0 1 6 14 21 0 4 6 9 12
Exercise:

Problem: Student grades on a chemistry exam were:
77, 78, 76, 81, 86, 51, 79, 82, 84, 99
e aConstruct a stem-and-leaf plot of the data.

e bAre there any potential outliers? If so, which scores are they?
Why do you consider them outliers?

Solution:

e b51,99

Try these multiple choice questions (Exercises 24 - 30).

The next three questions refer to the following information. We are
interested in the number of years students in a particular elementary
statistics class have lived in California. The information in the following
table is from the entire section.

Number of years Frequency

Number of years Frequency

7 |
14 3
15 1
18 1
19 4
20 3
22 1
23 1
26 1
AO 2
42 2
Total = 20
Exercise:

Problem: What is the IQR?

e A8

e Bll
e C15
e D35

Solution:

A

Exercise:

Problem: What is the mode?

e Al9

e B19.5

e C14 and 20
e D22.65

Solution:
A
Exercise:
Problem: Is this a sample or the entire population?

e Asample
e Bentire population
¢ Cneither

Solution:
B

The next two questions refer to the following table. X = the number of
days per week that 100 clients use a particular exercise facility.

x Frequency

0 3
1 12
2 33
3 28
4 11
5 9
6 4
Exercise:

Problem: The 80th percentile is:

e AS
e B80
e C3
e D4

Solution:

D
Exercise:

Problem:

The number that is 1.5 standard deviations BELOW the mean is
approximately:

e AO.7

e B48
e C-2.8
e DCannot be determined

Solution:

A

The next two questions refer to the following histogram. Suppose one
hundred eleven people who shopped in a special T-shirt store were asked
the number of T-shirts they own costing more than $19 each.

Relative
Frequency

40/111 39/111

30/111
25/111

23/111

20/111
17/111

10/111

Number of T-shirts costing more than $19 each

Exercise:

Problem:

The percent of people that own at most three (3) T-shirts costing more
than $19 each is approximately:

e A21
e B59
e C41
e DCannot be determined

Solution:

S
Exercise:
Problem:

If the data were collected by asking the first 111 people who entered
the store, then the type of sampling is:

e Acluster

¢ Bsimple random
e Cstratified

e Dconvenience

Solution:

D
Exercise:

Problem:

Below are the 2010 obesity rates by U.S. states and Washington,
DC.(Source: http://www.cdc.gov/obesity/data/adult.html))

State
Alabama
Alaska

Arizona

Arkansas

California
Colorado

Connecticut

Delaware

Washington,
DC

Florida
Georgia
Hawaii
Idaho

Illinois

Indiana

Percent
(%)

32.2
24.5

24.3

30.1

24.0
21.0

2225

28.0

22.2

26.6
29.6
22.7
26.5

28.2

29.6

State
Montana
Nebraska
Nevada

New
Hampshire

New Jersey
New Mexico
New York

North
Carolina

North
Dakota

Ohio
Oklahoma
Oregon
Pennsylvania
Rhode Island

South
Carolina

Percent
(%)

23.0
26.9

22.4

25.0

23.8
25.1

23.9

27.8

272

20.2
30.4
26.8
28.6

25:0

31.5

Percent Percent

State (%) State (%)
Iowa 28.4 =e 2753
Kansas 29.4 Tennessee 30.8
Kentucky 31.3 Texas 31.0
Louisiana 31.0 Utah 225
Maine 26.8 Vermont 2a
Maryland 27 Virginia 26.0
Massachusetts 23.0 Washington 25.5
Michigan 30.9 aah 32.5
Minnesota 24.8 Wisconsin 26.3
Mississippi 34.0 Wyoming 25.1
Missouri 30.5

e a.Construct a bar graph of obesity rates of your state and the four
States closest to your state. Hint: Label the x-axis with the states.

e b.Use a random number generator to randomly pick 8 states.
Construct a bar graph of the obesity rates of those 8 states.

e c.Construct a bar graph for all the states beginning with the letter
"AM

e d.Construct a bar graph for all the states beginning with the letter

Solution:

Example solution for b using the random number generator for the Ti-
84 Plus to generate a simple random sample of 8 states. Instructions
are below.

e Number the entries in the table 1 - 51 (Includes Washington, DC;
Numbered vertically)

e Press MATH

e Arrow over to PRB

e Press 5:randInt(

e Enter 51,1,8)

Eight numbers are generated (use the right arrow key to scroll through
the numbers). The numbers correspond to the numbered states (for this
example: {47 21 9 23 51 13 25 4}. If any numbers are repeated,
generate a different number by using 5:randInt(51,1)). Here, the states
(and Washington DC) are {Arkansas, Washington DC, Idaho,
Maryland, Michigan, Mississippi, Virginia, Wyoming}. Corresponding
percents are {28.7 21.8 24.5 26 28.9 32.8 25 24.6}.

40

35

30

25

Percent (%) 20
15

10

5

ie)
Arkansas WashDC Idaho Maryland Michigan Mississippi Virginia Wyoming

Exercise:

Problem:

A music school has budgeted to purchase 3 musical instruments. They
plan to purchase a piano costing $3000, a guitar costing $550, anda
drum set costing $600. The mean cost for a piano is $4,000 with a
standard deviation of $2,500. The mean cost for a guitar is $500 with a
standard deviation of $200. The mean cost for drums is $700 with a
standard deviation of $100. Which cost is the lowest, when compared
to other instruments of the same type? Which cost is the highest when
compared to other instruments of the same type. Justify your answer
numerically.

Solution:

For pianos, the cost of the piano is 0.4 standard deviations BELOW
the mean. For guitars, the cost of the guitar is 0.25 standard deviations
ABOVE the mean. For drums, the cost of the drum set is 1.0 standard
deviations BELOW the mean. Of the three, the drums cost the lowest
in comparison to the cost of other instruments of the same type. The
guitar cost the most in comparison to the cost of other instruments of
the same type.

Exercise:
Problem:
Suppose that a publisher conducted a survey asking adult consumers
the number of fiction paperback books they had purchased in the
previous month. The results are summarized in the table below. (Note

that this is the data presented for publisher B in homework exercise
13);

# of books Freq. Rel. Freq.

# of books Freq. Rel. Freq.

0 18

1 24

2 24

3 22

4 15

5 10

7 5

9 1

Publisher B
a. Are there any outliers in the data? Use an appropriate numerical

Lary

test involving the IQR to identify outliers, if any, and clearly state
your conclusion.

. If a data value is identified as an outlier, what should be done

about it?

. Are any data values further than 2 standard deviations away from

the mean? In some situations, statisticians may use this criteria to
identify data values that are unusual, compared to the other data
values. (Note that this criteria is most appropriate to use for data
that is mound-shaped and symmetric, rather than for skewed
data.)

. Do parts (a) and (c) of this problem give the same answer?
. Examine the shape of the data. Which part, (a) or (c), of this

question gives a more appropriate result for this data?

. Based on the shape of the data which is the most appropriate

measure of center for this data: mean, median or mode?

Solution:

e IQR=4-1=3; Q1—1.5*IQR = 1 —1.5(8) = -3.5; Q3 +
1.5*IQR = 4 + 1.5(3) = 8.5 ;The data value of 9 is larger than 8.5.
The purchase of 9 books in one month is an outlier.

e The outlier should be investigated to see if there is an error or
some other problem in the data; then a decision whether to
include or exclude it should be made based on the particular
situation. If it was a correct value then the data value should
remain in the data set. If there is a problem with this data value,
then it should be corrected or removed from the data. For
example: If the data was recorded incorrectly (perhaps a 9 was
miscoded and the correct value was 6) then the data should be
corrected. If it was an error but the correct value is not known it
should be removed from the data set.

e xbar— 2s = 2.45 — 2*1.88 = -1.31 ; xbar + 2s = 2.45 + 2*1.88 =
6.21 ; Using this method, the five data values of 7 books
purchased and the one data value of 9 books purchased would be
considered unusual.

e No: part (a) identifies only the value of 9 to be an outlier but part
(c) identifies both 7 and 9.

e The data is skewed (to the right). It would be more appropriate to
use the method involving the IQR in part (a), identifying only the
one value of 9 books purchased as an outlier. Note that part (c)
remarks that identifying unusual data values by using the criteria
of being further than 2 standard deviations away from the mean is
most appropriate when the data are mound-shaped and
symmetric.

e The data are skewed to the right. For skewed data it is more
appropriate to use the median as a measure of center.

**Fxercises 32 and 33 contributed by Roberta Bloom

Introduction to Bivariate Data

Measures of central tendency, variability, and spread summarize a single variable by providing
important information about its distribution. Often, more than one variable is collected on each
individual. For example, in large health studies of populations it is common to obtain variables
such as age, sex, height, weight, blood pressure, and total cholesterol on each individual.
Economic studies may be interested in, among other things, personal income and years of
education. As a third example, most university admissions committees ask for an applicant's high
school grade point average and standardized admission test scores (e.g., SAT). In this chapter we
consider bivariate data, which for now consists of two quantitative variables for each
individual. Our first interest is in summarizing such data in a way that is analogous to
summarizing univariate (single variable) data.

By way of illustration, let's consider something with which we are all familiar: age. It helps to
discuss something familiar since knowing the subject matter goes a long way in making
judgments about statistical results. Let's begin by asking if people tend to marry other people of
about the same age. Our experience tells us "yes," but how good is the correspondence? One way
to address the question is to look at pairs of ages for a sample of married couples. Table 1 below
shows the ages of 10 married couples. Going across the columns we see that, yes, husbands and
wives tend to be of about the same age, with men having a tendency to be slightly older than
their wives. This is no big surprise, but at least the data bear out our experiences, which is not
always the case.

Husband 36 72 37 36 51 50 47 50 37 41
Wife 35 67 3a 35 50 46 47 42 36 41
Sample of spousal ages of 10 White American Couples.
The pairs of ages in [link] are from a dataset consisting of 282 pairs of spousal ages, too many to
make sense of from a table. What we need is a way to summarize the 282 pairs of ages. We know

that each variable can be summarized by a histogram (see [link]) and by a mean and standard
deviation (See [link]).

Males Females

Histograms of spousal ages.

Mean Standard Deviation
Husband 49 11
Wife 47 11

Means and standard deviations of spousal ages.

Each distribution is fairly skewed with a long right tail. From [link] we see that not all husbands
are older than their wives and it is important to see that this fact is lost when we separate the
variables. That is, even though we provide summary statistics on each variable, the pairing within
couple is lost by separating the variables. We cannot say, for example, based on the means alone
what percentage of couples have younger husbands than wives. We have to count across pairs to
find this out. Only by maintaining the pairing can meaningful answers be found about couples
per se. Another example of information not available from the separate descriptions of husbands
and wives' ages is the mean age of husbands with wives of a certain age. For instance, what is the
average age of husbands with 45-year-old wives? Finally, we do not know the relationship
between the husband's age and the wife's age.

We can learn much more by displaying the bivariate data in a graphical form that maintains the
pairing. [link] shows a scatter plot of the paired ages. The x-axis represents the age of the
husband and the y-axis the age of the wife.

Wife's Age

Husband's Age

Scatterplot showing wife age as a function
of husband age.

There are two important characteristics of the data revealed by [link]. First, it is clear that there is
a strong relationship between the husband's age and the wife's age: the older the husband, the
older the wife. When one variable (y) increases with the second variable (v), we say that x and y
have a positive association. Conversely, when y decreases as zx increases, we say that they have
a Negative association.

Second, the points cluster along a straight line. When this occurs, the relationship is called a
linear relationship.

[link] shows a scatterplot of Arm Strength and Grip Strength from 149 individuals working in
physically demanding jobs including electricians, construction and maintenance workers, and
auto mechanics. Not surprisingly, the stronger someone's grip, the stronger their arm tends to be.
There is therefore a positive association between these variables. Although the points cluster
along a line, they are not clustered quite as closely as they are for the scatter plot of spousal age.

Arm Strength

Grip Strength

Scatter plot of Grip Strength and Arm
Strength.

Not all scatter plots show linear relationships. [link] shows the results of an experiment
conducted by Galileo on projectile motion. In the experiment, Galileo rolled balls down incline
and measured how far they traveled as a function of the release height. It is clear from [link] that
the relationship between "Release Height" and "Distance Traveled" is not described well by a
straight line: If you drew a line connecting the lowest point and the highest point, all of the
remaining points would be above the line. The data are better fit by a parabola.

600

500

400

Distance Traveled

300

0 250 500 750 1000 1250
Release Height

Galileo's data showing a non-linear
relationship.

Scatter plots that show linear relationships between variables can differ in several ways including
the slope of the line about which they cluster and how tightly the points cluster about the line. A
statistical measure of the strength of the relationship between variables that takes these factors
into account is the subject of the next section.

Glossary

Quantitative Variables
Variables that have are measured on a numeric or quantitative scale. Ordinal, interval and
ratio scales are quantitative. A country's population, a person's shoe size, or a car's speed are
all quantitative variables. Variables that are not quantitative are known as qualitative
variables.

Histogram
A histogram is a graphical representation of a distribution. It partitions the variable on the x-
axis into various contiguous class intervals of (usually) equal widths. The heights of the bars

represent the class frequencies.
150

_
NO
Oo

Frequency
8

395 495 595 695 795 895 995 1095 1195 1295 1395 1495 159.5 169.5

See also: Sturgis's Rule

Sturgis's Rule
One method of determining the number of classes for a histogram, Sturgis's Rule is to take
1 + log, N classes, rounded to the nearest integer.

Bivariate
Bivariate data is data for which there are two variables for each observation. As an example,
the following bivariate data show the ages of husbands and wives of 10 married couples.

Husband 36 72 a7 36 o1 50 47 50 a7 41

Wife ao 67 oo ao 50 46 47 42 36 41

Scatter Plot
A scatter plot of two variables shows the values of one variable on the Y axis and the values
of the other variable on the X axis. Scatter plots are well suited for revealing the relationship
between two variables. The scatter plot shown in [link] illustrates data from one of Galileo's
classic experiments in which he observed the distance traveled balls traveled after being
dropped on a incline as a function of their release height.

Positive Association
There is a positive association between variables X and Y if smaller values of X are
associated with smaller values of Y and larger values of X are assoicated with larger values
of Y.

Negative Association
There is a negative association between variables X and Y if smaller values of X are
associated with larger values of Y and larger values of X are assoicated with smaller values
of Y.

Linear Relationship
If the relationship between two variables is a perfect linear relationship, then a scatterplot of

the points will fall on a straight line as shown in [link].
20

With real data, there is almost never a perfect linear relationship between two variables. The
more the points tend to fall along a straight line the stronger the linear relationship. [link]

shows two variables (husband's age and wife's age) that have a strong but not a perfect
linear relationship.

Linear Regression and Correlation: Linear Equations

Linear regression for two variables is based on a linear equation with one
independent variable. It has the form:
Equation:

y=a+bx

where a and b are constant numbers.

x is the independent variable, and y is the dependent variable.
Typically, you choose a value to substitute for the independent variable and
then solve for the dependent variable.

Example:
The following examples are linear equations.
Equation:

v— 54 ok
Equation:

y = —0.01 + 1.2x

The graph of a linear equation of the form y = a + bx is a straight line.
Any line that is not vertical can be described by this equation.

Example:

15

Graph of the equation y = —1 + 2x.

Linear equations of this form occur in applications of life sciences, social

sciences, psychology, business, economics, physical sciences, mathematics,
and other areas.

Example:
Aaron's Word Processing Service (AWPS) does word processing. Its rate is
$32 per hour plus a $31.50 one-time charge. The total cost to a customer

depends on the number of hours it takes to do the word processing job.
Exercise:

Problem:

Find the equation that expresses the total cost in terms of the number
of hours required to finish the word processing job.

Solution:

Let x = the number of hours it takes to get the job done.
Let y = the total cost to the customer.

The $31.50 is a fixed cost. If it takes x hours to complete the job, then
(32) (a) is the cost of the word processing only. The total cost is:

y = 31.50 + 32x

Linear Regression and Correlation: Slope and Y-Intercept of a Linear
Equation

For the linear equation y = a + bx, b= slope and a = y-intercept.

From algebra recall that the slope is a number that describes the steepness
of a line and the y-intercept is the y coordinate of the point (0, a) where the
line crosses the y-axis.

If b > O, the line If b = 0, the If b < OQ, the line slopes
slopes upward to the line is downward to the right.
right. horizontal.

=| -—) FES

Three possible graphs of y = a + bx.

Example:

Svetlana tutors to make extra money for college. For each tutoring session,
she charges a one time fee of $25 plus $15 per hour of tutoring. A linear
equation that expresses the total amount of money Svetlana earns for each
session she tutors is y = 25 + 15x.

Exercise:

Problem:
What are the independent and dependent variables? What is the y-
intercept and what is the slope? Interpret them using complete

sentences.

Solution:

The independent variable (x) is the number of hours Svetlana tutors
each session. The dependent variable (y) is the amount, in dollars,
Svetlana earns for each session.

The y-intercept is 25 (a = 25). At the start of the tutoring session,
Svetlana charges a one-time fee of $25 (this is when x = 0). The slope
is 15 (b = 15). For each session, Svetlana earns $15 for each hour she
tutors.

The Regression Equation

Linear Regression and Correlation: The Regression Equation is a part of
Collaborative Statistics collection (col10522) by Barbara Illowsky and
Susan Dean. Contributions from Roberta Bloom include instructions for
finding and graphing the regression equation and scatterplot using the
LinRegTTest on the TI-83,83+,84+ calculators.

Data rarely fit a straight line exactly. Usually, you must be satisfied with
rough predictions. Typically, you have a set of data whose scatter plot
appears to "fit" a straight line. This is called a Line of Best Fit or Least
Squares Line.

Optional Collaborative Classroom Activity

If you know a person's pinky (smallest) finger length, do you think you
could predict that person's height? Collect data from your class (pinky
finger length, in inches). The independent variable, z, is pinky finger length
and the dependent variable, y, is height.

For each set of data, plot the points on graph paper. Make your graph big
enough and use a ruler. Then "by eye" draw a line that appears to "fit" the
data. For your line, pick two convenient points and use them to find the
slope of the line. Find the y-intercept of the line by extending your lines so
they cross the y-axis. Using the slopes and the y-intercepts, write your
equation of "best fit". Do you think everyone will have the same equation?
Why or why not?

Using your equation, what is the predicted height for a pinky length of 2.5
inches?

Example:

A random sample of 11 statistics students produced the following data
where z is the third exam score, out of 80, and y is the final exam score,
out of 200. Can you predict the final exam score of a random student if you
know the third exam score?

Table showing the scores on Scatter plot showing the scores

the final exam based on scores on the final exam based on scores
from the third exam. from the third exam.
250
8 200 - fe)
E 150 - 4 as eee
hy 100 -
y (final iz 7 |
x (third exam 60 65 70 75 80
exam score) score) Third Exam Score
65 75
67 188)
Ta 185
TEs 163
66 126
75 198
67 58)
70 163
7A 159
69 151

69 Sg

The third exam score, 2, is the independent variable and the final exam
score, y, is the dependent variable. We will plot a regression line that best
"fits" the data. If each of you were to fit a line "by eye", you would draw
different lines. We can use what is called a least-squares regression line to
obtain the best fit line.

Consider the following diagram. Each point of data is of the the form (a, y)
and each point of the line of best fit using least-squares linear regression has
the form (z, g).

The y is read "y hat" and is the estimated value of y. It is the value of y
obtained using the regression line. It is not generally equal to y from data.

data point = (x,, y,)

distance = ly, - y. =|e,

point on line = (x, y,)

64 69 74

The term yo — Yo = €o is called the "error" or residual. It is not an error
in the sense of a mistake. The absolute value of a residual measures the
vertical distance between the actual value of y and the estimated value of y.
In other words, it measures the vertical distance between the actual data
point and the predicted point on the line.

If the observed data point lies above the line, the residual is positive, and
the line underestimates the actual data value for y. If the observed data
point lies below the line, the residual is negative, and the line overestimates
that actual data value for y.

In the diagram above, yo — Yo = €o is the residual for the point shown. Here
the point lies above the line and the residual is positive.

€ = the Greek letter epsilon

For each data point, you can calculate the residuals or errors, y; — Y; = &;
fOr d= Ls 2. By aver Ld

Each |e| is a vertical distance.

For the example about the third exam scores and the final exam scores for
the 11 statistics students, there are 11 data points. Therefore, there are 11 €
values. If you square each € and add, you get

11
(2) (s)rr pe -be

This is called the Sum of Squared Errors (SSE).

Using calculus, you can determine the values of a and b that make the SSE
a minimum. When you make the SSE a minimum, you have determined the
points that are on the line of best fit. It turns out that the line of best fit has
the equation:

Equation:
y=a+bx
where a = y—b- x andb= oa

x and y are the sample means of the x values and the y values, respectively.
The best fit line always passes through the point (z, y).

The slope 6 can be written as b = r- (=) where s, = the standard

deviation of the y values and s,. = the standard deviation of the x values. r
is the correlation coefficient which is discussed in the next section.

Least Squares Criteria for Best Fit

The process of fitting the best fit line is called linear regression. The idea
behind finding the best fit line is based on the assumption that the data are
scattered about a straight line. The criteria for the best fit line is that the
sum of the squared errors (SSE) is minimized, that is made as small as
possible. Any other line you might choose would have a higher SSE than
the best fit line. This best fit line is called the least squares regression line

Note:Computer spreadsheets, statistical software, and many calculators
can quickly calculate the best fit line and create the graphs. The
calculations tend to be tedious if done by hand. Instructions to use the TI-
83, TI-83+, and TI-84+ calculators to find the best fit line and create a
scatterplot are shown at the end of this section.

THIRD EXAM vs FINAL EXAM EXAMPLE:
The graph of the line of best fit for the third exam/final exam example is
shown below:

250 -
200 -
i)
- :
Mm 150 -
= e ®
& 100 -
|
= 50
fe
0 i
64 69 74
Third Exam Score

The least squares regression line (best fit line) for the third exam/final exam
example has the equation:
Equation:

j = —-173.51 + 4.83x

Note:

e Remember, it is always important to plot a scatter diagram first. If the
scatter plot indicates that there is a linear relationship between the
variables, then it is reasonable to use a best fit line to make
predictions for y given x within the domain of z-values in the sample
data, but not necessarily for z-values outside that domain.

e You could use the line to predict the final exam score for a student
who eared a grade of 73 on the third exam.

e You should NOT use the line to predict the final exam score for a
student who earned a grade of 50 on the third exam, because 50 is not
within the domain of the x-values in the sample data, which are
between 65 and 75.

UNDERSTANDING SLOPE

The slope of the line, b, describes how changes in the variables are related.
It is important to interpret the slope of the line in the context of the situation
represented by the data. You should be able to write a sentence interpreting
the slope in plain English.

INTERPRETATION OF THE SLOPE: The slope of the best fit line tells
us how the dependent variable (y) changes for every one unit increase in the
independent (x) variable, on average.

THIRD EXAM vs FINAL EXAM EXAMPLE

e Slope: The slope of the line is b = 4.83.

e Interpretation: For a one point increase in the score on the third exam,
the final exam score increases by 4.83 points, on average.

Using the TI-83+ and TI-84+ Calculators
Using the Linear Regression T Test: LinRegT Test

In the STAT list editor, enter the X data in list L1 and the Y data in list L2,
paired so that the corresponding (x,y) values are next to each other in the
lists. (If a particular pair of values is repeated, enter it as many times as it
appears in the data.)

On the STAT TESTS menu, scroll down with the cursor to select the
LinRegT Test. (Be careful to select LinRegTTest as some calculators may
also have a different item called LinRegTInt.)

On the LinRegTTest input screen enter: Xlist: L1 ; Ylist: L2 ; Freq: 1

On the next line, at the prompt £ or p, highlight "4 0" and press ENTER
Leave the line for "RegEq:" blank

Highlight Calculate and press ENTER.

LinRegTTest Input Screen and Output Screen

LinRegT Test
Xlist: L1
Ylist: L2
Freq: 1

LinRegTTest
y=at+bx

£#0 and p#0
t= 2.657560155
p = .0261501512
df=9

B or p :[#0]<0 >0
RegEQ:
Calculate

Ja = -173.513363
b = 4.827394209

s = 16.41237711
r2= .4396931104
r= 663093591

TI-83+ and TI-84+
calculators

The output screen contains a lot of information. For now we will focus on a
few items from the output, and will return later to the other items.

e The second line says y=at+bx. Scroll down to find the values
a=-173.513, and b=4.8273 ; the equation of the best fit line is
y = —173.51+ 4.832

¢ The two items at the bottom are r? = .43969 and r=.663. For now, just
note where to find these values; we will discuss them in the next two
sections.

Graphing the Scatterplot and Regression Line

We are assuming your X data is already entered in list L1 and your Y data is
in list L2

Press 2nd STATPLOT ENTER to use Plot 1

On the input screen for PLOT 1, highlightOnand press ENTER

For TYPE: highlight the very first icon which is the scatterplot and press
ENTER

Indicate Xlist: L1 and Ylist: L2

For Mark: it does not matter which symbol you highlight.

Press the ZOOM key and then the number 9 (for menu item "ZoomStat") ;
the calculator will fit the window to the data

To graph the best fit line, press the "Y=" key and type the equation
-173.5+4.83X into equation Y1. (The X key is immediately left of the STAT
key). Press ZOOM 9 again to graph it.

Optional: If you want to change the viewing window, press the WINDOW
key. Enter your desired window using Xmin, Xmax, Ymin, Ymax

**With contributions from Roberta Bloom

Correlation Coefficient and Coefficient of Determination

Linear Regression and Correlation: The Correlation Coefficient and
Coefficient of Determination is a part of Collaborative Statistics collection
(col10522) by Barbara Illowsky and Susan Dean with contributions from
Roberta Bloom. The name has been changed from Correlation Coefficient.

The Correlation Coefficient r

Besides looking at the scatter plot and seeing that a line seems reasonable,
how can you tell if the line is a good predictor? Use the correlation
coefficient as another indicator (besides the scatterplot) of the strength of
the relationship between x and y.

The correlation coefficient, r, developed by Karl Pearson in the early
1900s, is a numerical measure of the strength of association between the
independent variable x and the dependent variable y.

The correlation coefficient is calculated as
Equation:

n Dar-y — (Be) (Zy)
in Sa? — (Ex) [n- Dy? — (Sy)?|

where n = the number of data points.

If you suspect a linear relationship between z and y, then r can measure
how strong the linear relationship is.
What the VALUE of r tells us:

e The value of r is always between -1 and +1: -1 <r <1.

e The size of the correlation r indicates the strength of the linear
relationship between z and y. Values of r close to -1 or to +1 indicate a
stronger linear relationship between x and y.

e If r=0 there is absolutely no linear relationship between x and y (no
linear correlation).

e If r = 1, there is perfect positive correlation. If 7 = —1, there is
perfect negative correlation. In both these cases, all of the original data
points lie on a straight line. Of course, in the real world, this will not
generally happen.

What the SIGN of r tells us

e A positive value of r means that when z increases, y tends to increase
and when « decreases, y tends to decrease (positive correlation).

e A negative value of r means that when z increases, y tends to decrease
and when z decreases, y tends to increase (negative correlation).

e The sign of r is the same as the sign of the slope, b, of the best fit line.

Note:Strong correlation does not suggest that 2 causes y or y causes x. We
say "correlation does not imply causation." For example, every person
who learned math in the 17th century is dead. However, learning math does
not necessarily cause death!

Positive Correlation

A scatter plot
showing data
with a
positive
correlation.
O0<r<l

Negative Correlation

A scatter plot
showing data
with a
negative
correlation.
—l<r<0

Zero Correlation

A scatter plot
showing data
with zero
correlation. r
=0

The formula for r looks formidable. However, computer spreadsheets,
statistical software, and many calculators can quickly calculate r. The
correlation coefficient r is the bottom item in the output screens for the

LinRegTTest on the TI-83, TI-83+, or TI-84+ calculator (see previous
section for instructions).

The Coefficient of Determination

r? is called the coefficient of determination. r? is the square of the

correlation coefficient , but is usually stated as a percent, rather than in
decimal form. r? has an interpretation in the context of the data:

e r”, when expressed as a percent, represents the percent of variation in
the dependent variable y that can be explained by variation in the
independent variable x using the regression (best fit) line.

e 1-r?, when expressed as a percent, represents the percent of variation
in y that is NOT explained by variation in x using the regression line.
This can be seen as the scattering of the observed data points about the
regression line.

Consider the third exam/final exam example introduced in the previous
section

e The line of best fit is: 7 = —173.51 + 4.83x

e The correlation coefficient is r = 0.6631

¢ The coefficient of determination is r? = 0.66317 = 0.4397

¢ Interpretation of r’ in the context of this example:

e Approximately 44% of the variation (0.4397 is approximately 0.44) in
the final exam grades can be explained by the variation in the grades
on the third exam, using the best fit regression line.

e Therefore approximately 56% of the variation (1 - 0.44 = 0.56) in the
final exam grades can NOT be explained by the variation in the grades
on the third exam, using the best fit regression line. (This is seen as the
scattering of the points about the line.)

**With contributions from Roberta Bloom.

Glossary

Coefficient of Correlation
A measure developed by Karl Pearson (early 1900s) that gives the
strength of association between the independent variable and the
dependent variable. The formula is:
Equation:

ndxy — Ol #)QOvy)
Jind 2? — (C2) Ov - Oly)
where n is the number of data points. The coefficient cannot be more

then 1 and less then -1. The closer the coefficient is to +1, the stronger
the evidence of a significant linear relationship between x and y.

—

Guessing Correlations Simulation

Begin by answering the questions, even if you have to guess. The first time
you answer the questions you will not be told whether you are correct or
not.

Once you have answered all the questions, answer them again using the
simulation to help you. This time you will get feedback about each
individual answer.

Show Simulation

This demonstration does not have questions associated with it. Use it to
improve your understanding of what scatter plots of data with different
values of r lool like.

General Instructions
This demonstration allows you to learn about Pearson's correlation by
viewing scatter plots with different values of Pearson's r. In each case, you

will have an opportunity to guess the correlation. With a little practice, you
should get pretty good at it.

Step by Step Instructions

Show Questions

Click the button next to the correlation that you think is present in the
scatter plot. You will find out if you were right or wrong. If you were
wrong, try another value. Then click the "New data" button.

Applet failed to run. No Java plug-in was found.

Summary

After looking at a series of scatterplots, you should have a pretty good feel
for what a certain value of r means.

Prediction

Linear Regression and Correlation: Prediction is a part of Collaborative
Statistics collection (col10522) by Barbara Illowsky and Susan Dean with
contributions from Roberta Bloom.

Recall the third exam/final exam example.

We examined the scatterplot and showed that the correlation coefficient is
significant. We found the equation of the best fit line for the final exam
grade as a function of the grade on the third exam. We can now use the least
squares regression line for prediction.

Suppose you want to estimate, or predict, the final exam score of statistics
students who received 73 on the third exam. The exam scores (z-values)
range from 65 to 75. Since 73 is between the x-values 65 and 75,
substitute x = 73 into the equation. Then:

Equation:

g = —173.51 + 4.83(73) = 179.08

We predict that statistic students who earn a grade of 73 on the third exam
will earn a grade of 179.08 on the final exam, on average.

Example:
Recall the third exam/final exam example.
Exercise:

Problem:

What would you predict the final exam score to be for a student who
scored a 66 on the third exam?

Solution:

145.27

Exercise:

Problem:

What would you predict the final exam score to be for a student who
scored a 90 on the third exam?

Solution:

The x values in the data are between 65 and 75. 90 is outside of the
domain of the observed x values in the data (independent variable), so
you cannot reliably predict the final exam score for this student. (Even
though it is possible to enter x into the equation and calculate a y
value, you should not do so!)

To really understand how unreliable the prediction can be outside of
the observed x values in the data, make the substitution x = 90 into the
equation.

j = —173.51 + 4.83(90) = 261.19

The final exam score is predicted to be 261.19. The largest the final
exam score can be is 200.

Note:The process of predicting inside of the observed x values in the
data is called interpolation. The process of predicting outside of the
observed x values in the data is called extrapolation.

**With contributions from Roberta Bloom

Linear Regression and Correlation: Summary
Bivariate Data: Each data point has two values. The form is
Line of Best Fit or Least Squares Line (LSL):

= independent variable; = dependent variable

Residual:
Correlation Coefficient r:

1. Used to determine whether a line of best fit is good for prediction.

2. Between -1 and 1 inclusive. The closer is to 1 or -1, the closer the
original points are to a straight line.

3.If is negative, the slope is negative. If is positive, the slope is
positive.

A. If , then the line is horizontal.

Sum of Squared Errors (SSE): The smaller the SSE, the better the
original set of points fits the line of best fit.

Outlier: A point that does not seem to fit the rest of the data.

Practice: Linear Regression

This module provides a practice of Linear Regression and Correlation as a
part of Collaborative Statistics collection (col10522) by Barbara Illowsky
and Susan Dean.

Student Learning Outcomes

e The student will evaluate bivariate data and determine if a line is an
appropriate fit to the data.

Given

Below are real data for the first two decades of AIDS reporting. (Source:
Centers for Disease Control and Prevention, National Center for HIV, STD,
and TB Prevention)

Year # AIDS cases diagnosed # AIDS deaths
Pre-1981 91 29

1981 319 121

1982 1,170 453

1983 3,076 1,482

1984 6,240 3,466

1985 11,776 6,878

1986 19,032 11,987

1987 28,564 16,162

1988 35,447 20,868
1989 42,674 2759 |
1990 48,634 31,335
1991 59,660 36,560
1992 78,530 41,055
1993 78,834 44,730
1994 71,874 49,095
1995 68,505 49,456
1996 Do,047 38,510
1997 47,149 20,736
1998 38,393 19,005
1999 25,174 18,454
2000 20,02 2 17,347
2001 25,643 17,402
2002 26,464 16,371
Total 802,118 489,093

Adults and Adolescents only, United States

Note:We will use the columns “year” and “# AIDS cases diagnosed” for
all questions unless otherwise stated.

Graphing
Graph “year” vs. “# AIDS cases diagnosed.” Plot the points on the graph

located below in the section titled "Plot" . Do not include pre-1981.
Label both axes with words. Scale both axes.

Data

Exercise:

Problem:
Enter your data into your calculator or computer. The pre-1981 data
should not be included. Why is that so?

Linear Equation

Write the linear equation below, rounding to 4 decimal places:

Note:For any prediction questions, the answers are calculated using the
least squares (best fit) line equation cited in the solution.

Exercise:

Problem: Calculate the following:

eaa—
a

* @ Corr: =
e dn =(# of pairs)

Solution:

© aa = -3,448,225

¢e bb= 1750

e ccorr. = 0.4526

eal = 22
Exercise:

Problem: equation: 7 =

Solution:

y = -3,448,225 +1750x

Solve

Exercise:

Problem: Solve.

e a When x = 1985, y =

¢ b When x = 1990, y =
Solution:

© a 25,525

e b34,275

Plot

Plot the 2 above points on the graph below. Then, connect the 2 points to
form the regression line.

Obtain the graph on your calculator or computer.

Discussion Questions

Look at the graph above.
Exercise:

Problem: Does the line seem to fit the data? Why or why not?

Exercise:

Problem: Do you think a linear fit is best? Why or why not?
Exercise:

Problem:

Hand draw a smooth curve on the graph above that shows the flow of

the data.

Exercise:

Problem:

What does the correlation imply about the relationship between time
(years) and the number of diagnosed AIDS cases reported in the U.S.?

Exercise:

Problem:

Why is “year” the independent variable and “# AIDS cases
diagnosed.” the dependent variable (instead of the reverse)?

Exercise:
Problem: Solve.
e aWhen x = 1970, 9 =:
e bWhy doesn’t this answer make sense?
Solution:

e a-/25

Homework

Linear Regression and Correlation: Homework is a part of Collaborative Statistics collection (col10522) by
Barbara Illowsky and Susan Dean.

Exercise:

Problem: For each situation below, state the independent variable and the dependent variable.

e aA study is done to determine if elderly drivers are involved in more motor vehicle fatalities than
all other drivers. The number of fatalities per 100,000 drivers is compared to the age of drivers.

e bA study is done to determine if the weekly grocery bill changes based on the number of family
members.

e cInsurance companies base life insurance premiums partially on the age of the applicant.

dUtility bills vary according to power consumption.

eA study is done to determine if a higher education reduces the crime rate in a population.

Solution:

e alndependent: Age; Dependent: Fatalities
e dIndependent: Power Consumption; Dependent: Utility

Note:For any prediction questions, the answers are calculated using the least squares (best fit) line equation
cited in the solution.

Exercise:
Problem:

Recently, the annual number of driver deaths per 100,000 for the selected age groups was as follows
(Source: http://

http://www.census.gov/compendia/statab/cats/transportation/motor_vehicle_accidents_and_fatalities.html):

Age Number of Driver Deaths per 100,000
16-19 38
20-24 36
25-34 24
35-54 20

55-74 18

Age Number of Driver Deaths per 100,000

75+ 28

aFor each age group, pick the midpoint of the interval for the x value. (For the 75+ group, use 80.)
bUsing “ages” as the independent variable and “Number of driver deaths per 100,000” as the
dependent variable, make a scatter plot of the data.

cCalculate the least squares (best—fit) line. Put the equation in the form of: 7 = a + bx

dFind the correlation coefficient. Is it significant?

ePick two ages and find the estimated fatality rates.

fUse the two points in (e) to plot the least squares line on your graph from (b).

gBased on the above data, is there a linear relationship between age of a driver and driver fatality
rate?

hWhat is the slope of the least squares (best-fit) line? Interpret the slope.

Exercise:

Problem:

The average number of people in a family that received welfare for various years is given below.
(Source: House Ways and Means Committee, Health and Human Services Department)

Year Welfare family size
1969 4.0
1973 3.6
1975 3.2
1979 3.0
1983 3.0
1988 3.0
1991 2.9

aUsing “year” as the independent variable and “welfare family size” as the dependent variable,
make a scatter plot of the data.

bCalculate the least squares line. Put the equation in the form of: 7 = a + bx

cFind the correlation coefficient. Is it significant?

dPick two years between 1969 and 1991 and find the estimated welfare family sizes.

eUse the two points in (d) to plot the least squares line on your graph from (b).

fBased on the above data, is there a linear relationship between the year and the average number of
people in a welfare family?

gUsing the least squares line, estimate the welfare family sizes for 1960 and 1995. Does the least
squares line give an accurate estimate for those years? Explain why or why not.

e hAre there any outliers in the above data?

e iWhat is the estimated average welfare family size for 1986? Does the least squares line give an
accurate estimate for that year? Explain why or why not.

jWhat is the slope of the least squares (best-fit) line? Interpret the slope.

Solution:

¢ by = 88.7206 — 0.04322

e c-0.8533, Yes

e gNo

hNo.

12.93, Yes

e jslope = -0.0432. As the year increases by one, the welfare family size tends to decrease by 0.0432
people.

Exercise:
Problem:
Use the AIDS data from the practice for this section, but this time use the columns “year #” and “# new
AIDS deaths in U.S.” Answer all of the questions from the practice again, using the new columns.
Exercise:
Problem:

The height (sidewalk to roof) of notable tall buildings in America is compared to the number of stories
of the building (beginning at street level). (Source: Microsoft Bookshelf)

Height (in feet) Stories
1050 57
428 28
362 26
529 AO
790 60
401 22
380 38

1454 110

Height (in feet) Stories
1127 100

700 46

e aUsing “stories” as the independent variable and “height” as the dependent variable, make a scatter
plot of the data.

e bDoes it appear from inspection that there is a relationship between the variables?

e cCalculate the least squares line. Put the equation in the form of: 7 = a + bx

dFind the correlation coefficient. Is it significant?

eFind the estimated heights for 32 stories and for 94 stories.

fUse the two points in (e) to plot the least squares line on your graph from (b).

gBased on the above data, is there a linear relationship between the number of stories in tall

buildings and the height of the buildings?

e hAre there any outliers in the above data? If so, which point(s)?

e iWhat is the estimated height of a building with 6 stories? Does the least squares line give an
accurate estimate of height? Explain why or why not.

e jBased on the least squares line, adding an extra story is predicted to add about how many feet to a

building?

kWhat is the slope of the least squares (best-fit) line? Interpret the slope.

Solution:

e bYes

e cy = 102.4287 + 11.75852

e d0.9436; yes

e e478.70 feet; 1207.73 feet

e gYes

hYes; (57,1050)

e 4172.98; No

e j11.7585 feet

kslope = 11.7585. As the number of stories increases by one, the height of the building tends to
increase by 11.7585 feet.

Exercise:

Problem:

Below is the life expectancy for an individual born in the United States in certain years. (Source:
National Center for Health Statistics)

Year of Birth Life Expectancy

1930 59.7

Year of Birth

1940

1950

1965

1973

1982

1987

1992

2010

Life Expectancy

62.9

e aDecide which variable should be the independent variable and which should be the dependent

variable.

bDraw a scatter plot of the ordered pairs.

cCalculate the least squares line. Put the equation in the form of: 7 = a + bx

dFind the correlation coefficient. Is it significant?

eFind the estimated life expectancy for an individual born in 1950 and for one born in 1982.
fWhy aren’t the answers to part (e) the values on the above chart that correspond to those years?
gUse the two points in (e) to plot the least squares line on your graph from (b).

hBased on the above data, is there a linear relationship between the year of birth and life
expectancy?

iAre there any outliers in the above data?

jUsing the least squares line, find the estimated life expectancy for an individual born in 1850.

Does the least squares line give an accurate estimate for that year? Explain why or why not.

Exercise:

Problem:

kWhat is the slope of the least squares (best-fit) line? Interpret the slope.

The percent of female wage and salary workers who are paid hourly rates is given below for the years
1979 - 1992. (Source: Bureau of Labor Statistics, U.S. Dept. of Labor)

Year

1979

1980

1981

Percent of workers paid hourly rates
61.2
60.7

61.3

Year Percent of workers paid hourly rates
1982 61.3
1983 61.8
1984 61.7
1985 61.8
1986 62.0
1987 62.7
1990 62.8
1992 62.9

e aUsing “year” as the independent variable and “percent” as the dependent variable, make a scatter
plot of the data.

bDoes it appear from inspection that there is a relationship between the variables? Why or why
not?

cCalculate the least squares line. Put the equation in the form of: 7 = a + bx

e dFind the correlation coefficient. Is it significant?

eFind the estimated percents for 1991 and 1988.
fUse the two points in (e) to plot the least squares line on your graph from (b).

e gBased on the above data, is there a linear relationship between the year and the percent of female
wage and salary earners who are paid hourly rates?
e hAre there any outliers in the above data?

iWhat is the estimated percent for the year 2050? Does the least squares line give an accurate

estimate for that year? Explain why or why not?

Solution:

bYes

c Y = —266.8863 + 0.1656z

d0.9448; Yes

e62.8233; 62.3265

h yes; (1987, 62.7)

172.5937; No

jslope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to

jWhat is the slope of the least squares (best-fit) line? Interpret the slope.

increase by 0.1656.

Exercise:

Problem:

The maximum discount value of the Entertainment® card for the “Fine Dining” section, Edition 10, for
various pages is given below.

Page number Maximum value ($)

4 16
14 19
25 15
32 17
43 19
57 15
72 16
85 15
90 17

aDecide which variable should be the independent variable and which should be the dependent
variable.

bDraw a scatter plot of the ordered pairs.

cCalculate the least squares line. Put the equation in the form of: 7 = a + bx

dFind the correlation coefficient. Is it significant?

eFind the estimated maximum values for the restaurants on page 10 and on page 70.

fUse the two points in (e) to plot the least squares line on your graph from (b).

gDoes it appear that the restaurants giving the maximum value are placed in the beginning of the
“Fine Dining” section? How did you arrive at your answer?

hSuppose that there were 200 pages of restaurants. What do you estimate to be the maximum value
for a restaurant listed on page 200?

ils the least squares line valid for page 200? Why or why not?

jWhat is the slope of the least squares (best-fit) line? Interpret the slope.

The next two questions refer to the following data: The cost of a leading liquid laundry detergent in
different sizes is given below.

Size (ounces) Cost ($) Cost per ounce
16 3.99
32 4.99
64 5.99

200

10.99

Exercise:

Problem:

aUsing “size” as the independent variable and “cost” as the dependent variable, make a scatter plot.
bDoes it appear from inspection that there is a relationship between the variables? Why or why
not?

cCalculate the least squares line. Put the equation in the form of: 7 = a + bx

dFind the correlation coefficient. Is it significant?

elf the laundry detergent were sold in a 40 ounce size, find the estimated cost.

ff the laundry detergent were sold in a 90 ounce size, find the estimated cost.

gUse the two points in (e) and (f) to plot the least squares line on your graph from (a).

hDoes it appear that a line is the best way to fit the data? Why or why not?

iAre there any outliers in the above data?

jis the least squares line valid for predicting what a 300 ounce size of the laundry detergent would
cost? Why or why not?

kWhat is the slope of the least squares (best-fit) line? Interpret the slope.

Solution:

bYes

cy = 3.5984 + 0.03712

d0.9986; Yes

e$5.08

£$6.93

iNo

jNot valid

kslope = 0.0371. As the number of ounces increases by one, the cost of liquid detergent tends to
increase by $0.0371 or is predicted to increase by $0.0371 (about 4 cents).

Exercise:

Problem:

aComplete the above table for the cost per ounce of the different sizes.

bUsing “Size” as the independent variable and “Cost per ounce” as the dependent variable, make a
scatter plot of the data.

cDoes it appear from inspection that there is a relationship between the variables? Why or why
not?

dCalculate the least squares line. Put the equation in the form of: y = a + bx

eFind the correlation coefficient. Is it significant?

ff the laundry detergent were sold in a 40 ounce size, find the estimated cost per ounce.

gif the laundry detergent were sold in a 90 ounce size, find the estimated cost per ounce.

hUse the two points in (f) and (g) to plot the least squares line on your graph from (b).

iDoes it appear that a line is the best way to fit the data? Why or why not?

jAre there any outliers in the above data?

kIs the least squares line valid for predicting what a 300 ounce size of the laundry detergent would
cost per ounce? Why or why not?

IWhat is the slope of the least squares (best-fit) line? Interpret the slope.

Exercise:

Problem:

According to flyer by a Prudential Insurance Company representative, the costs of approximate probate
fees and taxes for selected net taxable estates are as follows:

Net Taxable Estate ($) Approximate Probate Fees and Taxes ($)
600,000 30,000

750,000 92,500

1,000,000 203,000

1,500,000 438,000

2,000,000 688,000

2,900,000 1,037,000

3,000,000 1,350,000

aDecide which variable should be the independent variable and which should be the dependent
variable.

bMake a scatter plot of the data.

cDoes it appear from inspection that there is a relationship between the variables? Why or why
not?

dCalculate the least squares line. Put the equation in the form of: 7 = a + bx

eFind the correlation coefficient. Is it significant?

fFind the estimated total cost for a net taxable estate of $1,000,000. Find the cost for $2,500,000.
gUse the two points in (f) to plot the least squares line on your graph from (b).

hDoes it appear that a line is the best way to fit the data? Why or why not?

iAre there any outliers in the above data?

jBased on the above, what would be the probate fees and taxes for an estate that does not have any
assets?

kWhat is the slope of the least squares (best-fit) line? Interpret the slope.

Solution:

cYes

dy = —337,424.6478 + 0.54632

e0.9964; Yes

£$208,875.35; $1,028,325.35

hYes

iNo

kslope = 0.5463. As the net taxable estate increases by one dollar, the approximate probate fees and
taxes tend to increase by 0.5463 dollars (about 55 cents).

Exercise:

Problem: The following are advertised sale prices of color televisions at Anderson’s.

Size (inches) Sale Price ($)
9 147

20 197

27 297

31 447

35 1177

40 2177

60 2497

e aDecide which variable should be the independent variable and which should be the dependent
variable.

e bMake a scatter plot of the data.

¢ cDoes it appear from inspection that there is a relationship between the variables? Why or why

not?

dCalculate the least squares line. Put the equation in the form of: 7 = a + bx

eFind the correlation coefficient. Is it significant?

fFind the estimated sale price for a 32 inch television. Find the cost for a 50 inch television.

gUse the two points in (f) to plot the least squares line on your graph from (b).

hDoes it appear that a line is the best way to fit the data? Why or why not?

iAre there any outliers in the above data?

jWhat is the slope of the least squares (best-fit) line? Interpret the slope.

Exercise:

Problem: Below are the average heights for American boys. (Source: Physician’s Handbook, 1990)

Age (years) Height (cm)
birth 50.8

2 83.8

Age (years) Height (cm)

3 91.4
5 106.6
7 119.3
10 137.1
14 157.5

e aDecide which variable should be the independent variable and which should be the dependent
variable.

¢ bMake a scatter plot of the data.

e cDoes it appear from inspection that there is a relationship between the variables? Why or why
not?

e dCalculate the least squares line. Put the equation in the form of: 7 = a + bx

e eFind the correlation coefficient. Is it significant?

e fFind the estimated average height for a one year-old. Find the estimated average height for an

eleven year-old.

gUse the two points in (f) to plot the least squares line on your graph from (b).

hDoes it appear that a line is the best way to fit the data? Why or why not?

iAre there any outliers in the above data?

jUse the least squares line to estimate the average height for a sixty-two year-old man. Do you

think that your answer is reasonable? Why or why not?

e kWhat is the slope of the least squares (best-fit) line? Interpret the slope.

Solution:

e cYes

°e dy = 65.0876 + 7.0948z

e e0.9761; yes

£72.2 cm; 143.13 cm

e hYes

iNo

j505.0 cm; No

e kslope = 7.0948. As the age of an American boy increases by one year, the average height tends to
increase by 7.0948 cm.

Exercise:

Problem:

The following chart gives the gold medal times for every other Summer Olympics for the women’s 100
meter freestyle (swimming).

Year Time (seconds)

1912 82.2
1924 72.4
1932 66.8
1952 66.8
1960 61.2
1968 60.0
1976 55.65
1984 55.92
1992 54.64
2000 53.8
2008 53.1

e aDecide which variable should be the independent variable and which should be the dependent
variable.

¢ bMake a scatter plot of the data.

¢ cDoes it appear from inspection that there is a relationship between the variables? Why or why
not?

e dCalculate the least squares line. Put the equation in the form of: y = a + bx

e eFind the correlation coefficient. Is the decrease in times significant?

e fFind the estimated gold medal time for 1932. Find the estimated time for 1984.

e gWhy are the answers from (f) different from the chart values?

hUse the two points in (f) to plot the least squares line on your graph from (b).

iDoes it appear that a line is the best way to fit the data? Why or why not?

e jUse the least squares line to estimate the gold medal time for the next Summer Olympics. Do you
think that your answer is reasonable? Why or why not?

The next three questions use the following state information.

# letters in Year entered the Rank for entering the Area (square
State name Union Union miles)
Alabama 7 1819 22 52,423

Colorado 1876 38 104,100

# letters in Year entered the Rank for entering the Area (square

State name Union Union miles)
Hawaii 1959 50 10,932
Iowa 1846 29 56,276
Maryland 1788 7 12,407
Missouri 1821 24 69,709
ae 1787 3 8,722
Ohio 1803 17 44,828
aoe 13 1788 8 32,008
Utah 1896 45 84,904
Wisconsin 1848 30 65,499
Exercise:
Problem:

We are interested in whether or not the number of letters in a state name depends upon the year the state
entered the Union.

e aDecide which variable should be the independent variable and which should be the dependent
variable.

bMake a scatter plot of the data.

¢ cDoes it appear from inspection that there is a relationship between the variables? Why or why
not?

dCalculate the least squares line. Put the equation in the form of: 7 = a + bx

e eFind the correlation coefficient. What does it imply about the significance of the relationship?
e fFind the estimated number of letters (to the nearest integer) a state would have if it entered the
Union in 1900. Find the estimated number of letters a state would have if it entered the Union in
1940.

gUse the two points in (f) to plot the least squares line on your graph from (b).

e hDoes it appear that a line is the best way to fit the data? Why or why not?

iUse the least squares line to estimate the number of letters a new state that enters the Union this
year would have. Can the least squares line be used to predict it? Why or why not?

Solution:

e cNo

e dy = 47.03 — 0.0216
e e-0.4280

e f6;5

Exercise:

Problem:

We are interested in whether there is a relationship between the ranking of a state and the area of the
state.

aLet rank be the independent variable and area be the dependent variable.

bWhat do you think the scatter plot will look like? Make a scatter plot of the data.

cDoes it appear from inspection that there is a relationship between the variables? Why or why
not?

dCalculate the least squares line. Put the equation in the form of: 7 = a + bx

eFind the correlation coefficient. What does it imply about the significance of the relationship?
fFind the estimated areas for Alabama and for Colorado. Are they close to the actual areas?

gUse the two points in (f) to plot the least squares line on your graph from (b).

hDoes it appear that a line is the best way to fit the data? Why or why not?

iAre there any outliers?

jUse the least squares line to estimate the area of a new state that enters the Union. Can the least
squares line be used to predict it? Why or why not?

kDelete “Hawaii” and substitute “Alaska” for it. Alaska is the fortieth state with an area of 656,424
square miles.

ICalculate the new least squares line.

mFind the estimated area for Alabama. Is it closer to the actual area with this new least squares line
or with the previous one that included Hawaii? Why do you think that’s the case?

nDo you think that, in general, newer states are larger than the original states?

Exercise:

Problem:

We are interested in whether there is a relationship between the rank of a state and the year it entered the
Union.

aLet year be the independent variable and rank be the dependent variable.

bWhat do you think the scatter plot will look like? Make a scatter plot of the data.

cWhy must the relationship be positive between the variables?

dCalculate the least squares line. Put the equation in the form of: 7 = a + bx

eFind the correlation coefficient. What does it imply about the significance of the relationship?
fLet’s say a fifty-first state entered the union. Based upon the least squares line, when should that
have occurred?

gUsing the least squares line, how many states do we currently have?

hWhy isn’t the least squares line a good estimator for this year?

Solution:

dj = —480.5845 + 0.27482
e0.9553
£1934

Exercise:

Problem:

Below are the percents of the U.S. labor force (excluding self-employed and unemployed ) that are
members of a union. We are interested in whether the decrease is significant. (Source: Bureau of Labor
Statistics, U.S. Dept. of Labor)

Year Percent
1945 35.5
1950 31.5
1960 31.4
1970 27.3
1980 21.9
1993 15.8
2011 11.8

e aLet year be the independent variable and percent be the dependent variable.

¢ bWhat do you think the scatter plot will look like? Make a scatter plot of the data.

cWhy will the relationship between the variables be negative?

dCalculate the least squares line. Put the equation in the form of: 7 = a + bx

eFind the correlation coefficient. What does it imply about the significance of the relationship?
fBased on your answer to (e), do you think that the relationship can be said to be decreasing?
gif the trend continues, when will there no longer be any union members? Do you think that will
happen?

The next two questions refer to the following information: The data below reflects the 1991-92 Reunion
Class Giving. (Source: SUNY Albany alumni magazine)

Class Year Average Gift Total Giving
1922 41.67 125
1927 60.75 1,215

1932 83.82 3,772

Class Year Average Gift Total Giving

1937 87.84 5,710
1947 88.27 6,003
1952 76.14 5,254
1957 52.29 4,393
1962 57.80 4,451
1972 42.68 18,093
1976 49.39 22,473
1981 46.87 20,997
1986 37.03 12,590
Exercise:
Problem:

We will use the columns “class year” and “total giving” for all questions, unless otherwise stated.

aWhat do you think the scatter plot will look like? Make a scatter plot of the data.

bCalculate the least squares line. Put the equation in the form of: 7 = a + bx

cFind the correlation coefficient. What does it imply about the significance of the relationship?
dFor the class of 1930, predict the total class gift.

eFor the class of 1964, predict the total class gift.

fFor the class of 1850, predict the total class gift. Why doesn’t this value make any sense?

Solution:

b 7 = —569,770.2796 + 296.0351
c0.8302

d$1577.46

e$11,642.66

f-$22,105.34

Exercise:

Problem:

We will use the columns “class year” and “average gift” for all questions, unless otherwise stated.

aWhat do you think the scatter plot will look like? Make a scatter plot of the data.

bCalculate the least squares line. Put the equation in the form of: 7 = a + bx

cFind the correlation coefficient. What does it imply about the significance of the relationship?
dFor the class of 1930, predict the average class gift.

eFor the class of 1964, predict the average class gift.

fFor the class of 2010, predict the average class gift. Why doesn’t this value make any sense?

Exercise:

Problem:

We are interested in exploring the relationship between the weight of a vehicle and its fuel efficiency
(gasoline mileage). The data in the table show the weights, in pounds, and fuel efficiency, measured in
miles per gallon, for a sample of 12 vehicles.

Weight Fuel Efficiency
2715 24
2570 28
2610 29
2750 38
3000 25
3410 22
3640 20
3700 26
3880 21
3900 18
4060 18
4710 15

aGraph a scatterplot of the data.

bFind the correlation coefficient and determine if it is significant.

cFind the equation of the best fit line.

dWrite the sentence that interprets the meaning of the slope of the line in the context of the data.
eWhat percent of the variation in fuel efficiency is explained by the variation in the weight of the
vehicles, using the regression line? (State your answer in a complete sentence in the context of the
data.)

fAccurately graph the best fit line on your scatterplot.

gFor the vehicle that weights 3000 pounds, find the residual (y-yhat). Does the value predicted by
the line underestimate or overestimate the observed data value?

hidentify any outliers, using either the graphical or numerical procedure demonstrated in the
textbook.

iThe outlier is a hybrid car that runs on gasoline and electric technology, but all other vehicles in
the sample have engines that use gasoline only. Explain why it would be appropriate to remove the

outlier from the data in this situation. Remove the outlier from the sample data. Find the new
correlation coefficient, coefficient of determination, and best fit line.

e jCompare the correlation coefficients and coefficients of determination before and after removing
the outlier, and explain in complete sentences what these numbers indicate about how the model
has changed.

Solution:

br = -0.8, significant

cyhat = 48.4-0.00725x

dFor every one pound increase in weight, the fuel efficiency tends to decrease (or is predicted to
decrease) by 0.00725 miles per gallon. (For every one thousand pounds increase in weight, the fuel
efficiency tends to decrease by 7.25 miles per gallon.)

e64% of the variation in fuel efficiency is explained by the variation in weight using the regression
line.

e gyhat=48.4-0.00725(3000)=26.65 mpg. y-yhat=25-26.65=-1.65. Because yhat=26.5 is greater than
y=25, the line overestimates the observed fuel efficiency.

h(2750,38) is the outlier. Be sure you know how to justify it using the requested graphical or
numerical methods, not just by guessing.

e iyhat = 42.4-0.00578x

e jWithout outlier, r=-0.885, rsquare=0.76; with outlier, r=-0.8, rsquare=0.64. The new linear model
is a better fit, after the outlier is removed from the data, because the new correlation coefficient is
farther from 0 and the new coefficient of determination is larger.

Exercise:
Problem:
The four data sets below were created by statistician Francis Anscomb. They show why it is important

to examine the scatterplots for your data, in addition to finding the correlation coefficient, in order to
evaluate the appropriateness of fitting a linear model.

Set 1 Set 2 Set 3 Set 4

x y x y x y x y
10 8.04 10 9.14 10 7.46 8 6.58
8 6.95 8 8.14 8 6.77 8 5.76
13 7.58 13 8.74 13 12.74 8 7.71
9 8.81 9 8.77 9 7.11 8 8.84
11 8.33 11 9.26 11 7.81 8 8.47

14 9.96 14 8.10 14 8.84 8 7.04

4 4.26
12 10.84
7 4.82
5 5.68

6.13

3.10

9.13

7.26

4.74

6 6.08
4 5.39
12 8.15
7 6.42
5 5.73

19

a. For each data set, find the least squares regression line and the correlation coefficient. What did you
discover about the lines and values of r?

For each data set, create a scatter plot and graph the least squares regression line. Use the graphs to
answer the following questions:

e bFor which data set does it appear that a curve would be a more appropriate model than a line?
e cWhich data set has an influential point (point close to or on the line that greatly influences the

best fit line)?

e dWhich data set has an outlier (obviously visible on the scatter plot with best fit line graphed)?
e eWhich data set appears to be the most appropriate to model using the least squares regression

line?

Solution:

a. All four data sets have the same correlation coefficient r=0.816 and the same least squares regression

line yhat=3+0.5x

b. Set 2; c. Set 4; d. Set 3; e. Set 1

12 Y Data Set 1

ON AD

Y

Data Set2

Y

Data Set 3

14 Y Data Set4

Try these multiple choice questions

Exercise:

Problem: A correlation coefficient of -0.95 means there is a

e AStrong positive correlation
e¢ BWeak negative correlation

between the two variables.

e CStrong negative correlation
e DNo Correlation

Solution:

C
Exercise:

Problem:

According to the data reported by the New York State Department of Health regarding West Nile Virus
(http://www.health.state.ny.us/nysdoh/westnile/update/update.htm) for the years 2000-2008, the
least squares line equation for the number of reported dead birds (a) versus the number of human West
Nile virus cases (y) is y = —10.2638 + 0.04912. If the number of dead birds reported in a year is 732,
how many human cases of West Nile virus can be expected? r = 0.5490

e ANo prediction can be made.

e B19.6
e C15
e D38.1

Solution:

A

The next three questions refer to the following data: (showing the number of hurricanes by category to
directly strike the mainland U.S. each decade) obtained from www.nhc.noaa.gov/gifs/table6.gif A major
hurricane is one with a strength rating of 3, 4 or 5.

Decade

1941-1950

1951-1960

1961-1970

1971-1980

1981-1990

1991-2000

2001 — 2004

Exercise:

Total Number of Hurricanes

24

17

14

12

15

14

9

Number of Major Hurricanes
10

8

Problem:

Using only completed decades (1941 — 2000), calculate the least squares line for the number of major
hurricanes expected based upon the total number of hurricanes.

© Ay=—1.67r + 0.5
e By = 0.5x — 1.67
e Cy=0.942 — 1.67
© Dy=-2x+1
Solution:
B
Exercise:

Problem: The correlation coefficient is 0.942. Is this considered significant? Why or why not?

e ANo, because 0.942 is greater than the critical value of 0.707
e BYes, because 0.942 is greater than the critical value of 0.707
e CNo, because 0942 is greater than the critical value of 0.811

e DYes, because 0.942 is greater than the critical value of 0.811

Solution:

D
Exercise:
Problem:
The data for 2001-2004 show 9 hurricanes have hit the mainland United States. The line of best fit

predicts 2.83 major hurricanes to hit mainland U.S. Can the least squares line be used to make this
prediction?

e ANo, because 9 lies outside the independent variable values

¢ BYes, because, in fact, there have been 3 major hurricanes this decade

e¢ CNo, because 2.83 lies outside the dependent variable values

e DYes, because how else could we predict what is going to happen this decade.

Solution:

A

**Exercises 21 and 22 contributed by Roberta Bloom

Introduction
This module introduces the concept of Probability, the chance of an event
occurring.

Student Learning Outcomes
By the end of this chapter, the student should be able to:

e Understand and use the terminology of probability.

e Determine whether two events are mutually exclusive and whether two
events are independent.

¢ Calculate probabilities using the Addition Rules and Multiplication
Rules.

e Construct and interpret Contingency Tables.

e Construct and interpret Venn Diagrams (optional).

e Construct and interpret Tree Diagrams (optional).

Introduction

It is often necessary to "guess" about the outcome of an event in order to
make a decision. Politicians study polls to guess their likelihood of winning
an election. Teachers choose a particular course of study based on what they
think students can comprehend. Doctors choose the treatments needed for
various diseases based on their assessment of likely results. You may have
visited a casino where people play games chosen because of the belief that
the likelihood of winning is good. You may have chosen your course of
study based on the probable availability of jobs.

You have, more than likely, used probability. In fact, you probably have an
intuitive sense of probability. Probability deals with the chance of an event
occurring. Whenever you weigh the odds of whether or not to do your
homework or to study for an exam, you are using probability. In this
chapter, you will learn to solve probability problems using a systematic
approach.

Optional Collaborative Classroom Exercise

Your instructor will survey your class. Count the number of students in the
class today.

e Raise your hand if you have any change in your pocket or purse.
Record the number of raised hands.

e Raise your hand if you rode a bus within the past month. Record the
number of raised hands.

e Raise your hand if you answered "yes" to BOTH of the first two
questions. Record the number of raised hands.

Use the class data as estimates of the following probabilities. P(change)
means the probability that a randomly chosen person in your class has
change in his/her pocket or purse. P(bus) means the probability that a
randomly chosen person in your class rode a bus within the last month and
so on. Discuss your answers.

e Find P(change).

e Find P(bus).

e Find P(change and bus) Find the probability that a randomly chosen
student in your class has change in his/her pocket or purse and rode a
bus within the last month.

e Find P(change| bus) Find the probability that a randomly chosen
student has change given that he/she rode a bus within the last month.
Count all the students that rode a bus. From the group of students who
rode a bus, count those who have change. The probability is equal to
those who have change and rode a bus divided by those who rode a
bus.

Terminology

Probability: Terminology is part of the collection col10555 written by
Barbara Illowsky and Susan Dean defines key terms related to Probability
and has contributions from Roberta Bloom.

Probability is a measure that is associated with how certain we are of
outcomes of a particular experiment or activity. An experiment is a
planned operation carried out under controlled conditions. If the result is
not predetermined, then the experiment is said to be a chance experiment.
Flipping one fair coin twice is an example of an experiment.

The result of an experiment is called an outcome. A sample space is a set
of all possible outcomes. Three ways to represent a sample space are to list
the possible outcomes, to create a tree diagram, or to create a Venn diagram.
The uppercase letter S' is used to denote the sample space. For example, if
you flip one fair coin, S = {H, T} where H = heads and T = tails are the
outcomes.

An event is any combination of outcomes. Upper case letters like A and B
represent events. For example, if the experiment is to flip one fair coin,
event A might be getting at most one head. The probability of an event A is
written P(A).

The probability of any outcome is the long-term relative frequency of
that outcome. Probabilities are between 0 and 1, inclusive (includes 0 and
1 and all numbers between these values). P(A) = 0 means the event A can
never happen. P(A) = 1 means the event A always happens. P(A) = 0.5
means the event A is equally likely to occur or not to occur. For example, if
you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the
relative fequency of heads approaches 0.5 (the probability of heads).

Equally likely means that each outcome of an experiment occurs with
equal probability. For example, if you toss a fair, six-sided die, each face
(1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair
coin, a Head(H) and a Tail(T) are equally likely to occur. If you randomly
guess the answer to a true/false question on an exam, you are equally likely
to select a correct answer or an incorrect answer.

To calculate the probability of an event A when all outcomes in the
sample space are equally likely, count the number of outcomes for event
A and divide by the total number of outcomes in the sample space. For
example, if you toss a fair dime and a fair nickel, the sample space is
{HH, TH, HT, TT} where T = tails and H = heads. The sample space
has four outcomes. A = getting one head. There are two outcomes

{HT, TH}. P(A) =2.
Suppose you roll one fair six-sided die, with the numbers {1,2,3,4,5,6} on
its faces. Let event & = rolling a number that is at least 5. There are two
outcomes {5, 6}. P(E) =2. If you were to roll the die only a few times,

you would not be surprised if your observed results did not match the
probability. If you were to roll the die a very large number of times, you
would expect that, overall, 2/6 of the rolls would result in an outcome of "at
least 5". You would not expect exactly 2/6. The long-term relative
frequency of obtaining this result would approach the theoretical probability
of 2/6 as the number of repetitions grows larger and larger.

This important characteristic of probability experiments is the known as the
Law of Large Numbers: as the number of repetitions of an experiment is
increased, the relative frequency obtained in the experiment tends to
become closer and closer to the theoretical probability. Even though the
outcomes don't happen according to any set pattern or order, overall, the
long-term observed relative frequency will approach the theoretical
probability. (The word empirical is often used instead of the word
observed.) The Law of Large Numbers will be discussed again in Chapter 7.

It is important to realize that in many situations, the outcomes are not
equally likely. A coin or die may be unfair, or biased . Two math
professors in Europe had their statistics students test the Belgian 1 Euro
coin and discovered that in 250 trials, a head was obtained 56% of the time
and a tail was obtained 44% of the time. The data seem to show that the
coin is not a fair coin; more repetitions would be helpful to draw a more
accurate conclusion about such bias. Some dice may be biased. Look at the
dice in a game you have at home; the spots on each face are usually small
holes carved out and then painted to make the spots visible. Your dice may
or may not be biased; it is possible that the outcomes may be affected by the

slight weight differences due to the different numbers of holes in the faces.
Gambling casinos have a lot of money depending on outcomes from rolling
dice, so casino dice are made differently to eliminate bias. Casino dice have
flat faces; the holes are completely filled with paint having the same density
as the material that the dice are made out of so that each face is equally
likely to occur. Later in this chapter we will learn techniques to use to work
with probabilities for events that are not equally likely.

"OR" Event:

An outcome is in the event A OR B if the outcome is in A or is in B or is
in both A and B. For example, let A = {1, 2, 3, 4, 5} and

B= {4, 5,6, 7,8}. A OR B = (1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4
and 5 are NOT listed twice.

"AND" Event:

An outcome is in the event A AND B if the outcome is in both A and B at
the same time. For example, let A and B be {1, 2, 3, 4, 5} and

{4, 5, 6, 7, 8}, respectively. Then A AND B = {4, 5}.

The complement of event A is denoted A’ (read "A prime"). A’ consists of
all outcomes that are NOT in A. Notice that P(A) + P(A’) = 1. For
example, let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then,

A’ = {5, 6}. P(A) =4, P(A’) =2, and P(A) + P(A’) =4+2=1
The conditional probability of A given B is written P(A|B). P(A|B) is
the probability that event A will occur given that the event B has already
occurred. A conditional reduces the sample space. We calculate the
probability of A from the reduced sample space B. The formula to calculate
P(A|B) is

P(A|B)= SST a B)

where P(B) is greater than 0.
For example, suppose we toss one fair, six-sided die. The sample space

S = {1, 2, 3, 4, 5, 6}. Let A = face is 2 or 3 and B = face is even (2, 4, 6).
To calculate P(A|B), we count the number of outcomes 2 or 3 in the

sample space B = {2, 4, 6}. Then we divide that by the number of
outcomes in B (and not S).

We get the same result by using the formula. Remember that S has 6
outcomes.

P(A|B) =
P(A andB) __ (the number of outcomes that are 2 or 3 andeveninS)/6 1/6 1
P(B) (the number of outcomes that are even in S) / 6 ~ 3/6 3

Understanding Terminology and Symbols

It is important to read each problem carefully to think about and understand
what the events are. Understanding the wording is the first very important
step in solving probability problems. Reread the problem several times if
necessary. Clearly identify the event of interest. Determine whether there is
a condition stated in the wording that would indicate that the probability is
conditional; carefully identify the condition, if any.

Exercise:

Problem:

In a particular college class, there are male and female students. Some
students have long hair and some students have short hair. Write the
symbols for the probabilities of the events for parts (a) through (j)
below. (Note that you can't find numerical answers here. You were not
given enough information to find any probability values yet;
concentrate on understanding the symbols.)

e Let F be the event that a student is female.

e Let M be the event that a student is male.

e Let S be the event that a student has short hair.
e Let L be the event that a student has long hair.

e a The probability that a student does not have long hair.

¢ b The probability that a student is male or has short hair.

¢ c The probability that a student is a female and has long hair.

e d The probability that a student is male, given that the student has
long hair.

e e The probability that a student has long hair, given that the
student is male.

e f Of all the female students, the probability that a student has
short hair.

e g Of all students with long hair, the probability that a student is
female.

e h The probability that a student is female or has long hair.

e i The probability that a randomly selected student is a male
student with short hair.

e j The probability that a student is female.

Solution:

e a P(L')=P(S)
b P(M or S)
c P(F and L)
d P(M|L)

e P(L|M)

f P(S|F)

g P(F|L)

h P(F or L)
e i P(M andS)
PCP)

**With contributions from Roberta Bloom

Glossary

Conditional Probability
The likelihood that an event will occur given that another event has
already occurred.

Equally Likely
Each outcome of an experiment has the same probability.

Experiment
A planned activity carried out under controlled conditions.

Event
A subset in the set of all outcomes of an experiment. The set of all
outcomes of an experiment is called a sample space and denoted
usually by S. An event is any arbitrary subset in S. It can contain one
outcome, two outcomes, no outcomes (empty subset), the entire
sample space, etc. Standard notations for events are capital letters such
as A, B, CG, etc.

Outcome (observation)
A particular result of an experiment.

Probability
A number between 0 and 1, inclusive, that gives the likelihood that a
specific event will occur. The foundation of statistics is given by the
following 3 axioms (by A. N. Kolmogorov, 1930’s): Let S denote the
sample space and A and B are two events in S'. Then:

* 0 < P(A) <1;

e If A and B are any two mutually exclusive events, then
P(Aor B) = P(A) + P(B).

* P(S)=1.

Sample Space
The set of all possible outcomes of an experiment.

Independent and Mutually Exclusive Events

Probability: Independent and Mutually Exclusive Events is part of the
collection col10555 written by Barbara Illowsky and Susan Dean and
explains the concept of independent events, where the probability of event
A does not have any effect on the probability of event B, and mutually
exclusive events, where events A and B cannot occur at the same time. The
module has contributions from Roberta Bloom.

Independent and mutually exclusive do not mean the same thing.

Independent Events

Two events are independent if the following are true:

» P(A|B) = P(A)

Two events A and B are independent if the knowledge that one occurred
does not affect the chance the other occurs. For example, the outcomes of
two roles of a fair die are independent events. The outcome of the first roll
does not change the probability for the outcome of the second roll. To show
two events are independent, you must show only one of the above
conditions. If two events are NOT independent, then we say that they are
dependent.

Sampling may be done with replacement or without replacement.

¢ With replacement: If each member of a population is replaced after it
is picked, then that member has the possibility of being chosen more
than once. When sampling is done with replacement, then events are
considered to be independent, meaning the result of the first pick will
not change the probabilities for the second pick.

e Without replacement:: When sampling is done without replacement,
then each member of a population may be chosen only once. In this
case, the probabilities for the second pick are affected by the result of

the first pick. The events are considered to be dependent or not
independent.

If it is not known whether A and B are independent or dependent, assume
they are dependent until you can show otherwise.

Mutually Exclusive Events

A and B are mutually exclusive events if they cannot occur at the same
time. This means that A and B do not share any outcomes and
P(A AND B) = 0.

For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
Let A = {1, 2,3, 4,5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}.

A AND B = {4,5}. P(A AND B) = and is not equal to zero.
Therefore, A and B are not mutually exclusive. A and C do not have any
numbers in common so P(A AND C) = 0. Therefore, A and Care
mutually exclusive.

If it is not known whether A and B are mutually exclusive, assume they
are not until you can show otherwise.

The following examples illustrate these definitions and terms.

Example:

Flip two fair coins. (This is an experiment.)

The sample space is {HH, HT, TH, TT} where T = tails and H = heads.
The outcomes are HH, HT, TH, and TT. The outcomes HT and TH are
different. The HT means that the first coin showed heads and the second
coin showed tails. The ‘TH means that the first coin showed tails and the
second coin showed heads.

e Let A = the event of getting at most one tail. (At most one tail means
0 or 1 tail.) Then A can be written as {HH, HT, TH}. The outcome
HH shows 0 tails. HT and TH each show 1 tail.

¢ Let B= the event of getting all tails. B can be written as {TT}. B is
the complement of A. So, B = A’. Also,
P(A) + P(B) = P(A) + P(A’) = 1.

e The probabilities for A and for B are P(A) = + and P(B) = i.

e Let C = the event of getting all heads. C = {HH}. Since B = {TT},
P(B AND C) = 0. Band C are mutually exclusive. (B and C have
no members in common because you cannot have all tails and all
heads at the same time.)

e Let D = event of getting more than one tail. D = {TT}. P(D) = +

¢ Let & = event of getting a head on the first roll. (This implies you can
get either a head or tail on the second roll.) FE = {HT, HH}.
Ph).

e Find the probability of getting at least one (1 or 2) tail in two flips.
Let F' = event of getting at least one tail in two flips.

Faleeen (EUR AUS (Absit 20a)

Example:

Roll one fair 6-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event
A =a face is odd. Then A = {1, 3, 5}. Let event B =a face is even. Then
B = {2, 4, 6}.

e Find the complement of A, A’. The complement of A, A’, is B
because A and B together make up the sample space.
P(A) + P(B) = P(A) + P(A’) = 1. Also, P(A) = 3 and
P(B) = ¢

e Let event C = odd faces larger than 2. Then C' = {3, 5}. Let event D
= all even faces smaller than 5. Then D = {2,4}. P(C and D) = 0
because you cannot have an odd and even face at the same time.
Therefore, C' and D are mutually exclusive events.

e Let event F = all faces less than 5. H = {1, 2, 3, 4}.
Exercise:

Problem:

Are Cand E mutually exclusive events? (Answer yes or no.)
Why or why not?

Solution:

No. C = {3, 5} and E = {1, 2, 3, 4}. P(C AND E) = <. Tobe
mutually exclusive, P(C AND E) must be 0.

e Find P(C|A). This is a conditional. Recall that the event C’ is {3, 5}
and event A is {1, 3, 5}. To find P(C|A), find the probability of C
using the sample space A. You have reduced the sample space from
the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So,

P(C|A) = 4

Example:

Let event G = taking a math class. Let event H = taking a science class.
Then, G AND H = taking a math class and a science class. Suppose
P(G) = 0.6, P(H) = 0.5, and P(G AND H) = 0.3. Are G and H
independent?

If G and A are independent, then you must show ONE of the following:

» P(G|H) = P(G)

Note:The choice you make depends on the information you have. You
could choose any of the methods here because you have the necessary
information.

Exercise:

Problem: Show that P(G|H) = P(G).

Solution:

P(G AND H
P(G|H) = “Sa = 38 = 0.6 = PG)

Exercise:
Problem: Show P(G AND H) = P(G)- P(H).

Solution:

P(G)-P(H) = 0.6-0.5 = 0.3 = P(G AND H)

Since G and H are independent, then, knowing that a person is taking a
science class does not change the chance that he/she is taking math. If the
two events had not been independent (that is, they are dependent) then
knowing that a person is taking a science class would change the chance
he/she is taking math. For practice, show that P(H|G) = P(H) to show
that G and H are independent events.

Example:

In a box there are 3 red cards and 5 blue cards. The red cards are marked
with the numbers 1, 2, and 3, and the blue cards are marked with the
numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the
box (you cannot see into it) and draw one card.

Let R = red card is drawn, B = blue card is drawn, & = even-numbered
card is drawn.

The sample space S = R1, R2, R3, B1, B2, B3, B4, B5. S has 8
outcomes.

e2B Ch 3. P(B) = 2. P(R AND B) = 0. (You cannot draw one
card that is both red and blue.)

e P(E) = 3. (There are 3 even-numbered cards, R2, B2, and B4.)

¢ P(E|B) = 2. (There are 5 blue cards: B1, B2, B3, B4, and Bd. Out
of the blue cards, there are 2 even cards: B2 and B4.)

¢ P(B|E) = &. (There are 3 even-numbered cards: R2, B2, and B4.
Out of the even-numbered cards, 2 are blue: B2 and B4.)

e The events R and B are mutually exclusive because
P{R.AND B) = 0:

e Let G = card with a number greater than 3. G = {B4, B5}.
B(G) = 2. Let = blue card numbered between 1 and 4, inclusive.
H = {B1, B2, B3, B4}. P(G|H) = 4. (The only card in H that has
a number greater than 3 is B4.) Since = +, P(G) = P(G|H)
which means that G and H are independent.

Example:

In a particular college class, 60% of the students are female. 50 % of all
students in the class have long hair. 45% of the students are female and
have long hair. Of the female students, 75% have long hair. Let F be the
event that the student is female. Let L be the event that the student has long
hair. One student is picked randomly. Are the events of being female and
having long hair independent?

e The following probabilities are given in this example:
eR (R= 060 P(e) 40- 50

e P(F AND L) = 0.45

e P(L|F) = 0.75

Note:The choice you make depends on the information you have. You
could use the first or last condition on the list for this example. You do not
know P(F|L) yet, so you can not use the second condition.

Solution 1

Check whether P(F and L) = P(F)P(L): We are given that P(F and L) = 0.45
; but P(F)P(L) = (0.60)(0.50)= 0.30 The events of being female and having
long hair are not independent because P(F and L) does not equal P(F)P(L).
Solution 2

check whether P(L|F) equals P(L): We are given that P(L|F) = 0.75 but
P(L) = 0.50; they are not equal. The events of being female and having
long hair are not independent.

Interpretation of Results

The events of being female and having long hair are not independent;
knowing that a student is female changes the probability that a student has
long hair.

**Example 5 contributed by Roberta Bloom

Glossary

Independent Events
The occurrence of one event has no effect on the probability of the
occurrence of any other event. Events A and B are independent if one
of the following is true: (1). P(A|B) = P(A); (2) P(B|A) = P(B);
(3) P(Aand B) = P(A)P(B).

Mutually Exclusive
An observation cannot fall into more than one class (category). Being
in more than one category prevents being in a mutually exclusive
category.

Two Basic Rules of Probability
This module introduces the multiplication and addition rules used when calculating
probabilities.

The Multiplication Rule

If A and B are two events defined on a sample space, then:
P(A AND B) = P(B)- P(A|B).

This rule may also be written as : P(A|B)= ee ae
(The probability of A given B equals the probability of A and B divided by the
probability of B.)

If A and B are independent, then P(A|B) = P(A). Then
P(A AND B) = P(A|B) P(B) becomes P(A AND B) = P(A) P(B).

The Addition Rule

If A and B are defined on a sample space, then:
P(A OR B) = P(A) + P(B) — P(A AND B).

If A and B are mutually exclusive, then P(A AND B) = 0. Then
P(A OR B) = P(A) + P(B) — P(A AND B) becomes
P(A OR B) = P(A) + P(B).

Example:
Klaus is trying to choose where to go on vacation. His two choices are: A = New
Zealand and B = Alaska

¢ Klaus can only afford one vacation. The probability that he chooses A is
P(A) = 0.6 and the probability that he chooses B is P(B) = 0.35.

e P(A and B) = 0 because Klaus can only afford to take one vacation

e Therefore, the probability that he chooses either New Zealand or Alaska is
P(A OR B) = P(A) + P(B) = 0.6 + 0.35 = 0.95. Note that the probability
that he does not choose to go anywhere on vacation must be 0.05.

Example:

Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going
to attempt two goals in a row in the next game.

A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event
Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in
streaks. The probability that he makes the second goal GIVEN that he made the first
goal is 0.90.

Exercise:

Problem: What is the probability that he makes both goals?

Solution:

The problem is asking you to find P(A AND B) = P(B AND A). Since
P(B|A) = 0.90:
Equation:

P(B AND A) = P(BJA) P(A) = 0.90*0.65 = 0.585

Carlos makes the first and second goals with probability 0.585.
Exercise:

Problem:
What is the probability that Carlos makes either the first goal or the second goal?
Solution:

The problem is asking you to find P(A OR B).
Equation:

P(A ORB) = P(A) + P(B) — P(A AND B) = 0.65 + 0.65 — 0.585 = 0.715

Carlos makes either the first goal or the second goal with probability 0.715.
Exercise:

Problem: Are A and B independent?

Solution:

No, they are not, because P(B AND A) = 0.585.
Equation:

P(B) - P(A) = (0.65) - (0.65) = 0.423
Equation:

0.423 4 0.585 = P(B AND A)

So, P(B AND A) is not equal to P(B) - P(A).

Exercise:

Problem: Are A and B mutually exclusive?

Solution:
No, they are not because P(A and B) = 0.585.

To be mutually exclusive, P(A AND B) must equal 0.

Example:

A community swim team has 150 members. Seventy-five of the members are advanced
swimmers. Forty-seven of the members are intermediate swimmers. The remainder are
novice swimmers. Forty of the advanced swimmers practice 4 times a week. Thirty of
the intermediate swimmers practice 4 times a week. Ten of the novice swimmers
practice 4 times a week. Suppose one member of the swim team is randomly chosen.
Answer the questions (Verify the answers):

Exercise:

Problem: What is the probability that the member is a novice swimmer?

Solution:

28
150

Exercise:

Problem: What is the probability that the member practices 4 times a week?

Solution:

80
150

Exercise:
Problem:

What is the probability that the member is an advanced swimmer and practices 4
times a week?

Solution:

40
150

Exercise:
Problem:
What is the probability that a member is an advanced swimmer and an

intermediate swimmer? Are being an advanced swimmer and an intermediate
swimmer mutually exclusive? Why or why not?

Solution:

P(advanced AND intermediate) = 0, so these are mutually exclusive events.
A swimmer cannot be an advanced swimmer and an intermediate swimmer at the
same time.

Exercise:

Problem:

Are being a novice swimmer and practicing 4 times a week independent events?
Why or why not?

Solution:

No, these are not independent events.

Equation:

P(novice AND practices 4 times per week) = 0.0667
Equation:

P(novice) - P(practices 4 times per week) = 0.0996
Equation:

0.0667 ¢ 0.0996

Example:

Studies show that, if she lives to be 90, about 1 woman in 7 (approximately 14.3%) will
develop breast cancer. Suppose that of those women who develop breast cancer, a test is
negative 2% of the time. Also suppose that in the general population of women, the test
for breast cancer is negative about 85% of the time. Let B = woman develops breast
cancer and let NV = tests negative. Suppose one woman is selected at random.

Exercise:

Problem:

What is the probability that the woman develops breast cancer? What is the
probability that woman tests negative?

Solution:

P= 0143 SPN) 0:85
Exercise:

Problem:

Given that the woman has breast cancer, what is the probability that she tests
negative?

Solution:

P(N|B) = 0.02
Exercise:

Problem:
What is the probability that the woman has breast cancer AND tests negative?
Solution:

Pe AND N)— PB) PNB) = (0143) = (0:02) — 00029
Exercise:

Problem:
What is the probability that the woman has breast cancer or tests negative?

Solution:

P(B ORN) = P(B) + P(N) — P(B AND N) = 0.143 + 0.85 — 0.0029 = 0.9901

Exercise:

Problem: Are having breast cancer and testing negative independent events?

Solution:

No. P(N) = 0.85; P(N|B) = 0.02. So, P(N|B) does not equal P(N)

Exercise:

Problem: Are having breast cancer and testing negative mutually exclusive?

Solution:

No. P(B AND N) = 0.0029. For B and N to be mutually exclusive,
P(B AND N) must be 0.

Glossary

Independent Events
The occurrence of one event has no effect on the probability of the occurrence of
any other event. Events A and B are independent if one of the following is true: (1).
P(A|B) = P(A); (2) P(B|A) = P(B); (3) P(Aand B) = P(A)P(B).

Mutually Exclusive
An observation cannot fall into more than one class (category). Being in more than
one category prevents being in a mutually exclusive category.

Sample Space
The set of all possible outcomes of an experiment.

Contingency Tables
This module introduces the contingency table as a way of determining conditional
probabilities.

A contingency table provides a way of portraying data that can facilitate calculating
probabilities. The table helps in determining conditional probabilities quite easily. The
table displays sample values in relation to two different variables that may be dependent
or contingent on one another. Later on, we will use contingency tables again, but in
another manner. Contingincy tables provide a way of portraying data that can facilitate
calculating probabilities.

Example:
Suppose a study of speeding violations and drivers who use car phones produced the
following fictional data:

Speeding violation in No speeding violation in

the last year the last year Total
Cappel 25 280 305
user
een ces 45 405 450
phone user
Total 70 685 755

The total number of people in the sample is 755. The row totals are 305 and 450. The
column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.
Calculate the following probabilities using the table

Exercise:

Problem: P(person is a car phone user) =

Solution:

number of car phone users __ 305
total number in study — 755

Exercise:

Problem: P(person had no violation in the last year) =

Solution:

number that had no violation __ 685
total number in study 55

Exercise:

Problem:
P(person had no violation in the last year AND was a car phone user) =

Solution:

280
795

Exercise:

Problem:

P(person is a car phone user OR person had no violation in the last year) =

Solution:

(355 +755) “755 = 55
Exercise:

Problem:

P(person is a car phone user GIVEN person had a violation in the last year) =
Solution:

2. (The sample space is reduced to the number of persons who had a violation.)
Exercise:

Problem:
P(person had no violation last year GIVEN person was not a car phone user) =

Solution:

== (The sample space is reduced to the number of persons who were not car phone

users. )

Example:
The following table shows a random sample of 100 hikers and the areas of hiking
preferred:

The Near Lakes and On Mountain
Sex Coastline Streams Peaks Total
Female 18 16 ae: 45
Male — — 14 55
Total = 41

Hiking Area Preference

Exercise:

Problem: Complete the table.

Solution:
The Near Lakes and On Mountain
Sex Coastline Streams Peaks Total
Female 18 16 11 45
Male 16 25 14 55
Total 34 41 25 100

Hiking Area Preference

Exercise:

Problem:
Are the events "being female" and "preferring the coastline" independent events?
Let F' = being female and let C’' = preferring the coastline.

* aP(F AND C) =
¢ bP(F) - P(C) =

Are these two numbers the same? If they are, then F and C’ are independent. If they
are not, then F’ and C are not independent.

Solution:
¢ aP(F AND C) = == =0.18
¢ bP(F)-P(C) = > - 4a = 0.45 -0.34 = 0.153

P(F AND C) # P(F) - P(C), so the events F' and C are not independent.
Exercise:

Problem:

Find the probability that a person is male given that the person prefers hiking near

lakes and streams. Let M = being male and let L = prefers hiking near lakes and
streams.

e¢ aWhat word tells you this is a conditional?
¢ bFill in the blanks and calculate the probability: P(__|__) =
e cls the sample space for this problem all 100 hikers? If not, what is it?

Solution:
e aThe word 'given' tells you that this is a conditional.
e bP(M|L) = 2
e cNo, the sample space for this problem is 41.

Exercise:

Problem:

Find the probability that a person is female or prefers hiking on mountain peaks.
Let F = being female and let P = prefers mountain peaks.

aP(F) =
3 DE CE =
¢ cP(F AND P) =
e dTherefore, P(F OR P) =

Solution:
_ 45
e aP(F) = Ry
¢ bP(P) = Fae
e cP(F AND P) = a
° dP(FORP)= 3 +4-#=2
Example:

Muddy Mouse lives in a cage with 3 doors. If Muddy goes out the first door, the
probability that he gets caught by Alissa the cat is ~ and the probability he is not caught
is 2. If he goes out the second door, the probability he gets caught by Alissa is - and
the probability he is not caught is 3. The probability that Alissa catches Muddy coming
out of the third door is - and the probability she does not catch Muddy is S$. It is

equally likely that Muddy will choose any of the three doors so the probability of
choosing each door is =

Caught or Not Door One Door Two Door Three Total
1 1 i

Caught TH Tol ‘ei

Not Caught “= — 4

Caught or Not Door One Door Two Door Three

Total

Door Choice

° The first entry == = (+) (4) is P(Door One AND Caught).
¢ The entry = = () SS is P(Door One AND Not Caught).

Verify the remaining entries.
Exercise:

Problem:

Total

Complete the probability contingency table. Calculate the entries for the totals.

Verify that the lower-right corner entry is 1.

Solution:

Caught or Not Door One Door Two Door Three

1 1 il
Cau ght 15 TD ras
Not Caught = a 4
Total aa ~ =

Door Choice

Exercise:

Problem: What is the probability that Alissa does not catch Muddy?

Solution:

AL
60

Exercise:

Total

19
60

AL
60

1

Problem:

What is the probability that Muddy chooses Door One OR Door Two given that
Muddy is caught by Alissa?

Solution:

=e
19

Note: You could also do this problem by using a probability tree. See the Tree Diagrams
(Optional) section of this chapter for examples.

Glossary

Contingency Table
The method of displaying a frequency distribution as a table with rows and columns
to show how two variables may be dependent (contingent) upon each other. The
table provides an easy way to calculate conditional probabilities.

Venn Diagrams

This module introduces Venn diagrams as a method for solving some probability
problems. This module is included in the Elementary Statistics textbook/collection as an
optional lesson.

A Venn diagram is a picture that represents the outcomes of an experiment. It generally
consists of a box that represents the sample space S together with circles or ovals. The
circles or ovals represent events.

Example:

Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an
equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}.
Then A AND B = {6} and A ORB = (1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is
as follows:

Example:

Flip 2 fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then
A = {TT, TH} and B = {TT, HT}. Therefore, A AND B = {TT}.

AOR Ss == sb Haan Han

The sample space when you flip two fair coins is S = {HH, HT, TH, TT}. The
outcome HH is in neither A nor B. The Venn diagram is as follows:

S

Example:

Forty percent of the students at a local college belong to a club and 50% work part
time. Five percent of the students work part time and belong to a club. Draw a Venn
diagram showing the relationships. Let C’' = student belongs to a club and PT = student
works part time.

S
C and PT

_ «s

If a student is selected at random find

The probability that the student belongs to a club. P(C) = 0.40.

The probability that the student works part time. P(PT) = 0.50.

The probability that the student belongs to a club AND works part time.

P(C AND PT) = 0.05.

The probability that the student belongs to a club given that the student works part
time.

Equation:

P(CIPT) = P(CANDPT) _ 0.05 _ 9,
7 P(PT) — 0.50 °°

The probability that the student belongs to a club OR works part time.
Equation:

P(C OR PT) = P(C) + P(PT) — P(C AND PT) = 0.40 + 0.50 — 0.05 = 0.85

Glossary

Venn Diagram

The visual representation of a sample space and events in the form of circles or ovals
showing their intersections.

Tree Diagrams

This module introduces tree diagrams as a method for making some probability
problems easier to solve. This module is included in the Elementary Statistics
textbook/collection as an optional lesson.

A tree diagram is a special type of graph used to determine the outcomes of an
experiment. It consists of "branches" that are labeled with either frequencies or
probabilities. Tree diagrams can make some probability problems easier to visualize
and solve. The following example illustrates how to use a tree diagram.

Example:

In an urn, there are 11 balls. Three balls are red (A) and 8 balls are blue (B). Draw
two balls, one at a time, with replacement. "With replacement" means that you put
the first ball back in the urn before you select the second ball. The tree diagram using
frequencies that show all the possible outcomes follows.

24BR 24RB

Total = 64+ 24+ 244+ 9= 121

The first set of branches represents the first draw. The second set of branches
represents the second draw. Each of the outcomes is distinct. In fact, we can list each
red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and
B8. Then the 9 RR outcomes can be written as:

R1R1 R1R2 R1R3 R2R1 R2R2 R2R3 R3R1 R3R2 R3R3

The other outcomes are similar.

There are a total of 11 balls in the urn. Draw two balls, one at a time, and with
replacement. There are 11 - 11 = 121 outcomes, the size of the sample space.
Exercise:

Problem: List the 24 BR outcomes: B1R1, B1R2, B1R3, ...
Solution:
B1R1 B1R2 B1R3 B2R1 B2R2 B2R3 B3R1 B3R2 B3R3 B4R1 B4R2

B4R3 B5R1 B5R2 B5R3 B6R1 B6R2 B6R3 B7R1 B7R2 B7R3 B&R1
B8R2 B8&R3

Exercise:

Problem: Using the tree diagram, calculate P(RR).
Solution:

9 33 Oo —

Exercise:

Problem: Using the tree diagram, calculate P(RB OR BR).

Solution:

P(RBORBR)=2-4+2-4=2
Exercise:

Problem:

Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw).

Solution:

P(R on 1st draw AND B on 2nd draw) = P(RB) = 4-4 =

Exercise:

Problem:
Using the tree diagram, calculate P(R on 2nd draw given B on 1st draw).
Solution:

P(R on 2nd draw given B on Ist draw) = P(R on 2nd | Bon Ist) = 4 = 4

This problem is a conditional. The sample space has been reduced to those
outcomes that already have a blue on the first draw. There are 24 + 64 = 88
possible outcomes Ce BR and 64 BB). Twenty-four of the 88 possible

outcomes are BR. = = ea

Exercise:

Problem: Using the tree diagram, calculate P(BB).

Solution:

P(BB) = =
Exercise:

Problem:

Using the tree diagram, calculate
P(B on the 2nd draw given R on the first draw).

Solution:
P(B on 2nd draw | R on Ist draw) = --

There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB).
The sample space is then9 + 24 = 33. Twenty-four of the 33 outcomes have
B on the second draw. The probability is then _

Example:
An urn has 3 red marbles and 8 blue marbles in it. Draw two marbles, one at a time,
this time without replacement from the urn. "Without replacement" means that you
do not put the first ball back before you select the second ball. Below is a tree
diagram. The branches are labeled with probabilities instead of frequencies. The
numbers at the ends of the branches are ae by maui the numbers on the
two corresponding branches, for example

oa =o

1* Draw

56 24 24 6
110 110 110 110
BB BR RB RR

Total — 26t24+24+6 _ 110 _ 4

110 110

Note:If you draw a red on the first draw from the 3 red possibilities, there are 2 red
left to draw on the second draw. You do not put back or replace the first ball after
you have drawn it. You draw without replacement, so that on the second draw there
are 10 marbles left in the urn.

Calculate the following probabilities using the tree diagram.
Exercise:

Problem: P(RR) =

Solution:
3 2 6
P(RR) = 3 - ig = iw
Exercise:
Problem: Fill in the blanks:

ELSE OUR IB) sec ape te (ae

Solution:

P(RB or BR) = 2-84 (£)(2)= 4

Exercise:

Problem: P(R on 2d | B on 1st) =

Solution:

P(R on 2d|Bonist) = 4
Exercise:

Problem: Fill in the blanks:

P(R on 1st and B on 2nd) = P(RB) = (__)(__) = +

110
Solution:
P eR onelstandb on 2nd). — F(R) (3) (4) = 2.
Exercise:
Problem: P(BB) =
Solution:

P(BB) = = - +

Exercise:

Problem: P(B on 2nd | R on Ist) =
Solution:

There are 6 + 24 outcomes that have R on the first draw (6 RR and 24 RB).

The 6 and the 24 are frequencies. They are also the numerators of the fractions
acm and “. The sample space is no longer 110 but6 + 24 = 30. Twenty-
four of the 30 outcomes have B on the second draw. The probability is then a.

Did you get this answer?

If we are using probabilities, we can label the tree in the following general way.

PR and R)= P(RR)

R on 2nd | R on 1st
B on 2nd | R on 1st
R on 2nd | B on 1st
B on 2nd | B on 1st

SS”

) here means P
) here means P
) here means P
) here means P

EEN oN eae aa
~=_—"{$" ~~”

Glossary

Sample Space
The set of all possible outcomes of an experiment.

Tree Diagram
The useful visual representation of a sample space and events in the form of a
“tree” with branches marked by possible outcomes simultaneously with
associated probabilities (frequencies, relative frequencies).

Summary of Formulas

This module provides a review of the probability formulas, including the
definitions of independent, complementary, and mutually exclusive events
as well as the addition and multiplication rules.

Formula

Complement

If A and A’ are complements then P(A) + P(A’) = 1
Formula
Addition Rule

P(A OR B) = P(A) + P(B) — P(A AND B)
Formula
Mutually Exclusive

If A and B are mutually exclusive then P(A AND B) = 0; so
P(A ORB) = P(A) + P(B).

Formula

Multiplication Rule

« P(A AND B) = P(B)P(A[B)
¢ P(A AND B) = P(A)P(BJA)

Formula
Independence

If A and B are independent then:

* P(A|B) = P(A)
* P(BJA) = P(B)
¢ P(A AND B) = P(A)P(B)

Practice 1: Contingency Tables

This module provides the opportunity for students to apply what they've learned about probability to
solve a series of problems given a set of data. Students will practice constructing and interpreting
contingency tables.

Student Learning Outcomes

e The student will construct and interpret contingency tables.

Given

An article in the New England Journal of Medicine , reported about a study of smokers in California
and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were
given. Of the people smoking at most 10 cigarettes per day, there were 9886 African Americans, 2745
Native Hawaiians, 12,831 Latinos, 8378 Japanese Americans, and 7650 Whites. Of the people
smoking 11-20 cigarettes per day, there were 6514 African Americans, 3062 Native Hawaiians, 4932
Latinos, 10,680 Japanese Americans, and 9877 Whites. Of the people smoking 21-30 cigarettes per
day, there were 1671 African Americans, 1419 Native Hawaiians, 1406 Latinos, 4715 Japanese
Americans, and 6062 Whites. Of the people smoking at least 31 cigarettes per day, there were 759

African Americans, 788 Native Hawaiians, 800 Latinos, 2305 Japanese Americans, and 3970 Whites.
((Source: http://www.nejm.org/doi/full/10.1056/NEJMoa033250))

Complete the Table

Complete the table below using the data provided.

Smoking African Native Japanese
Level American Hawaiian Latino Americans White TOTALS

1-10
11-20
21-30
31+
TOTALS

Smoking Levels by Ethnicity

Analyze the Data

Suppose that one person from the study is randomly selected.

Exercise:

Problem: Find the probability that person smoked 11-20 cigarettes per day.

Solution:

35,065
100,450

Exercise:

Problem: Find the probability that person was Latino.

Solution:

19,969
100,450

Discussion Questions

Exercise:

Problem:

In words, explain what it means to pick one person from the study and that person is “Japanese
American AND smokes 21-30 cigarettes per day.” Also, find the probability.
Solution:

4,715
100,450

Exercise:

Problem:

In words, explain what it means to pick one person from the study and that person is “Japanese
American OR smokes 21-30 cigarettes per day.” Also, find the probability.

Solution:

36,636
100,450

Exercise:

Problem:

In words, explain what it means to pick one person from the study and that person is “Japanese
American GIVEN that person smokes 21-30 cigarettes per day.” Also, find the probability.

Solution:

4715
15,273

Exercise:

Problem: Prove that smoking level/day and ethnicity are dependent events.

Practice 2: Calculating Probabilities

This module allows students to practice using what they've learned about Probability. Students will apply their
understanding of basic probability terms, calculate probabilities based on the data provided, and determine whether
events are independent or mutually exclusive.

Student Learning Outcomes
e Students will define basic probability terms.

e Students will calculate probabilities.
e Students will determine whether two events are mutually exclusive or whether two events are independent.

Note: Use probability rules to solve the problems below. Show your work.

Given

48% of all Californians registered voters prefer life in prison without parole over the death penalty for a person
convicted of first degree murder. Among Latino California registered voters, 55% prefer life in prison without
parole over the death penalty for a person convicted of first degree murder. (Source:
http://field.com/fieldpollonline/subscribers/RIs2393.pdf ).

37.6% of all Californians are Latino (Source: U.S. Census Bureau).

In this problem, let:

C = Californians (registered voters) preferring life in prison without parole over the death penalty f
e L = Latino Californians

Suppose that one Californian is randomly selected.

Analyze the Data
Exercise:
Problem: P(C) =
Solution:

0.48

Exercise:

Problem: P(L) =

Solution:

0.376

Exercise:
Problem: P(C|L) =

Solution:

0.55

Exercise:

Problem: In words, what is " C'|L"?

Exercise:

Problem: P(L AND C) =

Solution:

0.2068

Exercise:

Problem: [In words, what is “Z and C”’?

Exercise:

Problem: Are L and C’ independent events? Show why or why not.

Solution:
No

Exercise:

Problem: P(L OR C) =

Solution:
0.6492

Exercise:

Problem: [In words, what is “Z or C”?

Exercise:

Problem: Are LZ and C’' mutually exclusive events? Show why or why not.

Solution:

No

Homework

Probability: Homework is part of the collection col10555 written by Barbara Illowsky and Susan
Dean and provides a number of homework exercises related to Probability with contributions
from Roberta Bloom.

Exercise:

Problem:

Suppose that you have 8 cards. 5 are green and 3 are yellow. The 5 green cards are
numbered 1, 2, 3, 4, and 5. The 3 yellow cards are numbered 1, 2, and 3. The cards are well
shuffled. You randomly draw one card.

e G=card drawn is green
e K=card drawn is even-numbered

e aList the sample space.

* bP(G) =

¢ cP(G|E) =

e dP(G AND £) =

¢ eP(G ORE) =

e fAre G and F mutually exclusive? Justify your answer numerically.

Solution:

e a{G1, G2, G3, G4, G5, Y1, Y2, Y3}
° b2

e c2
e dz
e es

e {No

Exercise:

Problem:

Refer to the previous problem. Suppose that this time you randomly draw two cards, one at a
time, and with replacement.

¢ G,= first card is green
e G»= second card is green

e aDraw a tree diagram of the situation.
eb P(G, AND G2) =

¢ c P(at least one green) =

¢ d P(Gy | Gi) =

e eAre G2 and G; independent events? Explain why or why not.

Exercise:

Problem:

Refer to the previous problems. Suppose that this time you randomly draw two cards, one at
a time, and without replacement.

G = first card is green
G 9= second card is green

aDraw a tree diagram of the situation.

bP(G, AND Ga) =

cP(at least one green) =

dP(G2|G1) =

eAre Gz» and G, independent events? Explain why or why not.

Solution:
© b(3)(F)
© (33) + (s)(F) + (B))
e d 4
e eNo
Exercise:

Problem: Roll two fair dice. Each die has 6 faces.

aList the sample space.

bLet A be the event that either a 3 or 4 is rolled first, followed by an even number. Find
P(A).

cLet B be the event that the sum of the two rolls is at most 7. Find P(B).

din words, explain what “P(A|B)” represents. Find P(A|B).

eAre A and B mutually exclusive events? Explain your answer in 1 - 3 complete
sentences, including numerical justification.

fAre A and B independent events? Explain your answer in 1 - 3 complete sentences,
including numerical justification.

Exercise:

Problem:

A special deck of cards has 10 cards. Four are green, three are blue, and three are red. When
a card is picked, the color of it is recorded. An experiment consists of first picking a card
and then tossing a coin.

e aList the sample space.

¢ bLet A be the event that a blue card is picked first, followed by landing a head on the
coin toss. Find P(A).

¢ cLet B be the event that a red or green is picked, followed by landing a head on the
coin toss. Are the events A and B mutually exclusive? Explain your answer in 1 - 3
complete sentences, including numerical justification.

e dLet C’ be the event that a red or blue is picked, followed by landing a head on the coin
toss. Are the events A and C’ mutually exclusive? Explain your answer in 1 - 3
complete sentences, including numerical justification.

Solution:

¢ a{GH,GT,BH,BT,RH,RT}

Exercise:

Problem: An experiment consists of first rolling a die and then tossing a coin:

e aList the sample space.

¢ bLet A be the event that either a 3 or 4 is rolled first, followed by landing a head on the
coin toss. Find P(A).

¢ cLet B be the event that a number less than 2 is rolled, followed by landing a head on
the coin toss. Are the events A and B mutually exclusive? Explain your answer in 1 - 3
complete sentences, including numerical justification.

Exercise:

Problem:

An experiment consists of tossing a nickel, a dime and a quarter. Of interest is the side the
coin lands on.

e aList the sample space.

¢ bLet A be the event that there are at least two tails. Find P(A).

e cLet B be the event that the first and second tosses land on heads. Are the events A and
B mutually exclusive? Explain your answer in 1 - 3 complete sentences, including
justification.

Solution:

- a{(HHH),(HHT),(HTH),(HTT),(THH),(THT),(TTH),(TTT)}
e b<
e cYes

Exercise:

Problem: Consider the following scenario:

¢ Let P(C) = 0.4
* Let P(D) = 0.5
* Let P(C|D) = 0.6

¢ aFind P(C AND D).

bAre C’ and D mutually exclusive? Why or why not?
cAre C and D independent events? Why or why not?
dFind P(C ORD).

eFind P(D|C).

Exercise:

Problem: E£ and F mutually exclusive events. P(E) = 0.4; P(F) = 0.5. Find P(E | F).

Solution:

0

Exercise:

Problem: J and K are independent events. P(J | K) = 0.3. Find P(J) .

Exercise:

Problem: U and V are mutually exclusive events. P(U) = 0.26; P(V) = 0.37. Find:

« aP(U AND V) =
* bP(U| V) =
cP(U OR V) =

Solution:

e a0
« bo
e c0.63
Exercise:
Problem:
Q and R are independent events. P(Q) = 0.4; P(Q AND R) = 0.1. Find P(R).

Exercise:

Problem: Y and Z are independent events.

¢ a Rewrite the basic Addition Rule P(Y OR Z) = P(Y) + P(Z) — P(Y AND Z)
using the information that Y and Z are independent events.
¢ b Use the rewritten rule to find P(Z) if P(Y OR Z) = 0.71 and P(Y) = 0.42.

Solution:

e b0.5
Exercise:

Problem: G and H are mutually exclusive events. P(G) = 0.5; P(H) = 0.3

¢ aExplain why the following statement MUST be false: P(H | G) = 0.4.
¢ bFind: P(H OR G).
e cAre G and H independent or dependent events? Explain in a complete sentence.

Exercise:
Problem:
The following are real data from Santa Clara County, CA. As of a certain time, there had

been a total of 3059 documented cases of AIDS in the county. They were grouped into the
following categories (Source: Santa Clara County Public H.D.):

IV
Drug Heterosexual
Homosexual/Bisexual User* Contact Other Totals
Female 0 70 136 49
Male 2146 463 60 135

Totals a
* includes homosexual/bisexual IV drug users

Suppose one of the persons with AIDS in Santa Clara County is randomly selected.
Compute the following:

¢ a P(person is female) =

¢ bP(person has a risk factor Heterosexual Contact) =

¢ cP(person is female OR has a risk factor of IV Drug User) =
e dP(person is female AND has a risk factor of Homosexual/Bisexual) =
¢ eP(person is male AND has a risk factor of IV Drug User) =
¢ fP(female GIVEN person got the disease from heterosexual contact) =

e gConstruct a Venn Diagram. Make one group females and the other group heterosexual

contact.

Solution:

The completed contingency table is as follows:

IV
Drug
Homosexual/Bisexual User*
Female 0 70
Male 2146 463
Totals 2146 533

* includes homosexual/bisexual IV drug users

e
aa
a
w
aD

Exercise:

Problem:

Heterosexual
Contact

136

60

196

Other

49

135

184

Totals

255

2804

3059

Solve these questions using probability rules. Do NOT use the contingency table above.
3059 cases of AIDS had been reported in Santa Clara County, CA, through a certain date.

Those cases will be our population. Of those cases, 6.4% obtained the disease through

heterosexual contact and 7.4% are female. Out of the females with the disease, 53.3% got

the disease from heterosexual contact.

* aP(person is female) =

¢ bP(person obtained the disease through heterosexual contact) =

¢ cP(female GIVEN person got the disease from heterosexual contact) =

e dConstruct a Venn Diagram. Make one group females and the other group heterosexual
contact. Fill in all values as probabilities.

Exercise:

Problem:

The following table identifies a group of children by one of four hair colors, and by type of
hair.

Hair Type Brown Blond Black Red Totals
Wavy 20 15 3 43
Straight 80 15 12

Totals 20 215

¢ aComplete the table above.

e¢ bWhat is the probability that a randomly selected child will have wavy hair?

e cWhat is the probability that a randomly selected child will have either brown or blond
hair?

e¢ dWhat is the probability that a randomly selected child will have wavy brown hair?

e eWhat is the probability that a randomly selected child will have red hair, given that he
has straight hair?

e fif B is the event of a child having brown hair, find the probability of the complement
of B.

e gin words, what does the complement of B represent?

Solution:

Exercise:

Problem:

A previous year, the weights of the members of the San Francisco 49ers and the Dallas
Cowboys were published in the San Jose Mercury News. The factual data are compiled into
the following table.

Shirt# < 210 211-250 251-290 290<
1-33 21 iS) 0 0
34-66 6 18 Z 4
66-99 6 12 22 5

For the following, suppose that you randomly select one player from the 49ers or Cowboys.

e aFind the probability that his shirt number is from 1 to 33.

e bFind the probability that he weighs at most 210 pounds.

e cFind the probability that his shirt number is from 1 to 33 AND he weighs at most 210
pounds.

e dFind the probability that his shirt number is from 1 to 33 OR he weighs at most 210
pounds.

e eFind the probability that his shirt number is from 1 to 33 GIVEN that he weighs at
most 210 pounds.

e flf having a shirt number from 1 to 33 and weighing at most 210 pounds were
independent events, then what should be true about P(Shirt# 1-33 | < 210 pounds)?

Exercise:

Problem:

Approximately 281,000,000 people over age 5 live in the United States. Of these people,
55,000,000 speak a language other than English at home. Of those who speak another
language at home, 62.3% speak Spanish. (Source:
http://www.census.gov/hhes/socdemo/language/data/acs/ACS-12.pdf)

Let: F = speak English at home; E’ = speak another language at home; S = speak Spanish;

Finish each probability statement by matching the correct answer.

Probability Statements
a. P(E’) =

b. P(E) =

c. P(S and E') =

d. P(SIE') =

Solution:

e aiil

e bi

e civ

e dii
Exercise:

Problem:

Answers

i. 0.8043

ii. 0.623

iii. 0.1957

iv. 0.1219

The probability that a male develops some form of cancer in his lifetime is 0.4567 (Source:
American Cancer Society). The probability that a male has at least one false positive test
result (meaning the test comes back for cancer when the man does not have it) is 0.51
(Source: USA Today). Some of the questions below do not have enough information for you
to answer them. Write “not enough information” for those answers.

Let: C' = a man develops cancer in his lifetime; P = man has at least one false positive

e aConstruct a tree diagram of the situation.

* bP(C) =
¢ cP(P|C) =
¢ dP(P|C’ ) =

e elf atest comes up positive, based upon numerical values, can you assume that man has

cancer? Justify numerically and explain why or why not.

Exercise:

Problem:

In 1994, the U.S. government held a lottery to issue 55,000 Green Cards (permits for non-
citizens to work legally in the U.S.). Renate Deutsch, from Germany, was one of
approximately 6.5 million people who entered this lottery. Let G = won Green Card.

¢ aWhat was Renate’s chance of winning a Green Card? Write your answer as a

probability statement.

e bin the summer of 1994, Renate received a letter stating she was one of 110,000
finalists chosen. Once the finalists were chosen, assuming that each finalist had an
equal chance to win, what was Renate’s chance of winning a Green Card? Let
F = was a finalist. Write your answer as a conditional probability statement.

e cAre G and F independent or dependent events? Justify your answer numerically and
also explain why.

e dAre G and F mutually exclusive events? Justify your answer numerically and also
explain why.

Note:P.S. Amazingly, on 2/1/95, Renate learned that she would receive her Green Card --
true story!

Solution:

¢ a P(G) = 0.008
e bO.5

¢ cdependent

e dNo

Exercise:

Problem:

Three professors at George Washington University did an experiment to determine if
economists are more selfish than other people. They dropped 64 stamped, addressed
envelopes with $10 cash in different classrooms on the George Washington campus. 44%
were returned overall. From the economics classes 56% of the envelopes were returned.
From the business, psychology, and history classes 31% were returned. (Source: Wall Street
Journal)

Let: R = money returned; EF’ = economics classes; O = other classes

e aWrite a probability statement for the overall percent of money returned.

e bWrite a probability statement for the percent of money returned out of the economics
classes.

e cWrite a probability statement for the percent of money returned out of the other
classes.

e dis money being returned independent of the class? Justify your answer numerically
and explain it.

e eBased upon this study, do you think that economists are more selfish than other
people? Explain why or why not. Include numbers to justify your answer.

Exercise:

Problem:

The chart below gives the number of suicides estimated in the U.S. for a recent year by age,
race (black and white), and sex. We are interested in possible relationships between age,
race, and sex. We will let suicide victims be our population. (Source: The National Center
for Health Statistics, U.S. Dept. of Health and Human Services)

Race and Sex 1-14 15 - 24 25 - 64 over 64 TOTALS
white, male 210 3360 13,610 22,050
white, female 80 580 3380 4930
black, male 10 460 1060 1670
black, female 0 40 270 330

all others

TOTALS 310 4650 18,780 29,760

Note:Do not include "all others" for parts (f), (g), and (i).

aFill in the column for the suicides for individuals over age 64.

bFill in the row for all other races.

cFind the probability that a randomly selected individual was a white male.

dFind the probability that a randomly selected individual was a black female.

eFind the probability that a randomly selected individual was black

fFind the probability that a randomly selected individual was male.

gOut of the individuals over age 64, find the probability that a randomly selected
individual was a black or white male.

hComparing “Race and Sex” to “Age,” which two groups are mutually exclusive? How
do you know?

iAre being male and committing suicide over age 64 independent events? How do you
know?

Solution:

22050
29760
330

29760
2000
29760

f 23720

ie

8 6020
¢ hBlack females and ages 1-14
e iNo

The next two questions refer to the following: The percent of licensed U.S. drivers (from a
recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age
20 - 64; 13.61% are age 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and
under; 81.43% are age 20 - 64; 13.53% are age 65 or over. (Source: Federal Highway
Administration, U.S. Dept. of Transportation)

Exercise:

Problem: Complete the following:

e aConstruct a table or a tree diagram of the situation.

¢ bP(driver is female) =

¢ cP(driver is age 65 or over | driver is female) =

e dP(driver is age 65 or over AND female) =

e elIn words, explain the difference between the probabilities in part (c) and part (d).
¢ fP(driver is age 65 or over) =

e gAre being age 65 or over and being female mutually exclusive events? How do you
know

Exercise:

Problem: Suppose that 10,000 U.S. licensed drivers are randomly selected.

e aHow many would you expect to be male?

e bUsing the table or tree diagram from the previous exercise, construct a contingency
table of gender versus age group.

e cUsing the contingency table, find the probability that out of the age 20 - 64 group, a
randomly selected driver is female.

Solution:

e a5140
e c0.49

Exercise:

Problem:

Approximately 86.5% of Americans commute to work by car, truck or van. Out of that
group, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work
and approximately 5.3% take public transportation. (Source: Bureau of the Census, U.S.
Dept. of Commerce. Disregard rounding approximations.)

e aConstruct a table or a tree diagram of the situation. Include a branch for all other
modes of transportation to work.

¢ bAssuming that the walkers walk alone, what percent of all commuters travel alone to
work?

¢ cSuppose that 1000 workers are randomly selected. How many would you expect to
travel alone to work?

e dSuppose that 1000 workers are randomly selected. How many would you expect to
drive in a carpool?

Exercise:

Problem: Explain what is wrong with the following statements. Use complete sentences.

¢ alf there’s a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then
there’s a 130% chance of rain over the weekend.

e bThe probability that a baseball player hits a home run is greater than the probability
that he gets a successful hit.

Try these multiple choice questions.

The next two questions refer to the following probability tree diagram which shows tossing
an unfair coin FOLLOWED BY drawing one bead from a cup containing 3 red (R), 4 yellow (Y
) and 5 blue (B) beads. For the coin, P(H) = 2 and P(T') = + where H = ”heads” and

T=" tails”.

R 3/12
Yan
2/3
H SB sn2
R
: 3/12
1/3
Y 4/12

Exercise:

Problem: Find P(tossing a Head on the coin AND a Red bead)

e A
° B
°C
e D

w w loreal Wb
slagjaalee

Solution:

C

Exercise:

Problem: Find P(Blue bead).

eA
- Be
ec 2
ops

Solution:
A
The next three questions refer to the following table of data obtained from www.baseball-

almanac.com showing hit information for 4 well known baseball players. Suppose that one hit
from the table is randomly selected.

NAME Single Double Triple Home Run TOTAL HITS
Babe Ruth 1517 506 136 714 2873
Jackie Robinson 1054 273 54 137 1518
Ty Cobb 3603 174 295 114 4189

Hank Aaron 2294 624 98 759 3771

NAME Single Double Triple

TOTAL 8471 1577 583
Exercise:

Problem: Find P(hit was made by Babe Ruth).

° A ian
oe
© D a5357

Solution:

B

Exercise:

Home Run TOTAL HITS

1720 12351

Problem: Find P(hit was made by Ty Cobb | The hit was a Home Run)

Solution:

B
Exercise:

Problem:

Are the hit being made by Hank Aaron and the hit being a double independent

events?

¢ A Yes, because P(hit by Hank Aaron | hit is a double) = P(hit by Hank Aaron)
¢ BNo, because P(hit by Hank Aaron | hit is a double) 4 P(hit is a double)

e C No, because

P(hit is by Hank Aaron | hit is a double) ¢ P(hit by Hank Aaron)
¢ D Yes, because P(hit is by Hank Aaron | hit is a double) = P(hit is a double)

Solution:

Cc

Exercise:

Problem: Given events G and H: P(G) = 0.43 ; P(H) = 0.26 ; P(H and G) = 0.14

e A Find P(H or G)
e BFind the probability of the complement of event (H and G)
e CFind the probability of the complement of event (H or G)

Solution:

¢ A P(H orG) = P(H) + P(G) - P(H and G) = 0.26 + 0.43 - 0.14 = 0.55
¢ BP( NOT (H andG) ) = 1 - P(HandG) = 1-0.14=0.86
* C P( NOT (HorG)) =1- P(HorG)=1-0.55=0.45

Exercise:

Problem: Given events J and K: P(J) = 0.18 ; P(K) = 0.37 ; PJ or K) = 0.45

e A Find P(J and k)
e B Find the probability of the complement of event (J and K)
¢ C Find the probability of the complement of event (J or K)

Solution:

e AP(J or K) = P(J) + P(K) - PU and K); 0.45 = 0.18 + 0.37 — PJ and K) ; solve to find
P(J and K) = 0.10

¢ BP( NOT (J and K) ) = 1 —- PV and K) = 1 - 0.10 = 0.90

e C P( NOT (J or K) ) =1- PU or K) =1-0.45=0.55

Exercise:

Problem:

United Blood Services is a blood bank that serves more than 500 hospitals in 18 states.
According to their website, http://www.unitedbloodservices.org/humanbloodtypes.html, a
person with type O blood and a negative Rh factor (Rh-) can donate blood to any person
with any bloodtype. Their data show that 43% of people have type O blood and 15% of
people have Rh- factor; 52% of people have type O or Rh- factor.

e A Find the probability that a person has both type O blood and the Rh- factor
e B Find the probability that a person does NOT have both type O blood and the Rh-
factor.

Solution:

e A P(Type O or Rh-) = P(Type O) + P(Rh-) — P(Type O and Rh-)
0.52 = 0.43 + 0.15 — P(Type O and Rh-); solve to find P(Type O and Rh-) = 0.06
6% of people have type O Rh- blood

¢ BP( NOT (Type O and Rh-) ) = 1 — P(Type O and Rh-) = 1 — 0.06 = 0.94
94% of people do not have type O Rh- blood

Exercise:
Problem:
At a college, 72% of courses have final exams and 46% of courses require research papers.

Suppose that 32% of courses have a research paper and a final exam. Let F be the event that
a course has a final exam. Let R be the event that a course requires a research paper.

e A Find the probability that a course has a final exam or a research project.
e B Find the probability that a course has NEITHER of these two requirements.

Solution:

e A P(R or F) = P(R) + P(F) — P(R and F) = 0.72 + 0.46 — 0.32 = 0.86
¢ B P( Neither R nor F ) = 1 - P(Ror F) =1-0.86=0.14

Exercise:
Problem:

In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. Of those, 8%
contain both chocolate and nuts. Sean is allergic to both chocolate and nuts.

e A Find the probability that a cookie contains chocolate or nuts (he can't eat it).
e B Find the probability that a cookie does not contain chocolate or nuts (he can eat it).

Solution:

e Let C be the event that the cookie contains chocolate. Let N be the event that the cookie
contains nuts.

e A P(C or N) = P(C) + P(N) — P(C and N) = 0.36 + 0.12 — 0.08 = 0.40

¢ B P( neither chocolate nor nuts) = 1 — P(C or N) = 1 - 0.40 = 0.60

Exercise:

Problem:

A college finds that 10% of students have taken a distance learning class and that 40% of
students are part time students. Of the part time students, 20% have taken a distance learning
class. Let D = event that a student takes a distance learning class and E = event that a student
is a part time student

A Find P(D and E)

B Find P(E | D)

C Find P(D or E)

¢ D Using an appropriate test, show whether D and E are independent.

e E Using an appropriate test, show whether D and E are mutually exclusive.

Solution:

¢ A P(D and E) = P(D|E)P(E) = (0.20)(0.40) = 0.08

B P(E|D) = P(D and E) / P(D) = 0.08/0.10 = 0.80

C P(D or E) = P(D) + P(E) — P(D and E) = 0.10 + 0.40 — 0.08 = 0.42

¢ D Not Independent: P(D|E) = 0.20 which does not equal P(D) = .10

e E Not Mutually Exclusive: P(D and E) = 0.08 ; if they were mutually exclusive then we
would need to have P(D and E) = 0, which is not true here.

Exercise:

Problem:

When the Euro coin was introduced in 2002, two math professors had their statistics
students test whether the Belgian 1 Euro coin was a fair coin. They spun the coin rather than
tossing it, and it was found that out of 250 spins, 140 showed a head (event H) while 110
showed a tail (event T). Therefore, they claim that this is not a fair coin.

e A Based on the data above, find P(H) and P(T).

¢ B Use a tree to find the probabilities of each possible outcome for the experiment of
tossing the coin twice.

e C Use the tree to find the probability of obtaining exactly one head in two tosses of the
coin.

¢ D Use the tree to find the probability of obtaining at least one head.

Solution:

¢ A P(H) = 140/250; P(T) = 110/250
° €308/625
° D 504/625

Exercise:
Problem:
A box of cookies contains 3 chocolate and 7 butter cookies. Miguel randomly selects a

cookie and eats it. Then he randomly selects another cookie and eats it also. (How many
cookies did he take?)

e A Draw the tree that represents the possibilities for the cookie selections. Write the
probabilities along each branch of the tree.

¢ B Are the probabilities for the flavor of the SECOND cookie that Miguel selects
independent of his first selection? Explain.

e C For each complete path through the tree, write the event it represents and find the
probabilities.

¢ DLet S be the event that both cookies selected were the same flavor. Find P(S).

e E Let T be the event that both cookies selected were different flavors. Find P(T) by two
different methods: by using the complement rule and by using the branches of the tree.
Your answers should be the same with both methods.

e F Let U be the event that the second cookie selected is a butter cookie. Find P(U).

**Exercises 33 - 40 contributed by Roberta Bloom

Review
This module provides a number of homework/review exercises related to
Probability.

The first six exercises refer to the following study: In a survey of 100
stocks on NASDAQ, the average percent increase for the past year was 9%
for NASDAQ stocks. Answer the following:

Exercise:

Problem: The “average increase” for all NASDAQ stocks is the:
e APopulation
e BStatistic
e CParameter
¢ DSample
e EVariable

Solution:

e C Parameter
Exercise:

Problem: All of the NASDAQ stocks are the:

e APopulation
e BStatistic
e CParameter
¢ DSample
e EVariable

Solution:

e A Population

Exercise:

Problem: 9% is the:
e APopulation
e BStatistic
e CParameter
¢ DSample
e EVariable

Solution:

e B Statistic
Exercise:

Problem: The 100 NASDAQ stocks in the survey are the:
e APopulation
e BStatistic
e CParameter
¢ DSample
e EVariable

Solution:

¢ D Sample
Exercise:

Problem: The percent increase for one stock in the survey is the:

e APopulation
e BStatistic

e CParameter
¢ DSample
e EVariable
Solution:
e E Variable
Exercise:

Problem:

Would the data collected be qualitative, quantitative — discrete, or
quantitative — continuous?

Solution:

quantitative - continuous

The next two questions refer to the following study: Thirty people spent
two weeks around Mardi Gras in New Orleans. Their two-week weight gain
is below. (Note: a loss is shown by a negative weight gain.)

Weight Gain Frequency
-2 3
-1 is)

0 2

Weight Gain Frequency

1 4

4 13

6 2

11 1
Exercise:

Problem: Calculate the following values:
e aThe average weight gain for the two weeks
e bThe standard deviation
e cThe first, second, and third quartiles
Solution:
e a2.27

° b3.04
° c-1,4,4

Exercise:

Problem: Construct a histogram and a boxplot of the data.

Introduction
This module serves as the introduction to Discrete Random Variables in the
Elementary Statistics textbook/collection.

Student Learning Outcomes
By the end of this chapter, the student should be able to:

¢ Recognize and understand discrete probability distribution functions,
in general.

e Calculate and interpret expected values.

e Recognize the binomial probability distribution and apply it
appropriately.

e Recognize the Poisson probability distribution and apply it
appropriately (optional).

e Recognize the geometric probability distribution and apply it
appropriately (optional).

e Recognize the hypergeometric probability distribution and apply it
appropriately (optional).

e Classify discrete word problems by their distributions.

Introduction

A student takes a 10 question true-false quiz. Because the student had such
a busy schedule, he or she could not study and randomly guesses at each
answer. What is the probability of the student passing the test with at least a
70%?

Small companies might be interested in the number of long distance phone
calls their employees make during the peak time of the day. Suppose the
average is 20 calls. What is the probability that the employees make more
than 20 long distance phone calls during the peak time?

These two examples illustrate two different types of probability problems
involving discrete random variables. Recall that discrete data are data that
you can count. A random variable describes the outcomes of a statistical

experiment in words. The values of a random variable can vary with each
repetition of an experiment.

In this chapter, you will study probability problems involving discrete
random distributions. You will also study long-term averages associated
with them.

Random Variable Notation

Upper case letters like X or Y denote a random variable. Lower case letters
like x or y denote the value of a random variable. If X is a random
variable, then X is written in words. and z is given as a number.

For example, let X = the number of heads you get when you toss three fair
coins. The sample space for the toss of three fair coins is TTT THH HTH
HHT HTT THT TTH HHH. Then, zx = 0, 1, 2, 3. X is in words and z is
a number. Notice that for this example, the x values are countable
outcomes. Because you can count the possible values that X can take on
and the outcomes are random (the z values 0, 1, 2, 3), X is a discrete
random variable.

Optional Collaborative Classroom Activity

Toss a coin 10 times and record the number of heads. After all members of
the class have completed the experiment (tossed a coin 10 times and
counted the number of heads), fill in the chart using a heading like the one
below. Let X = the number of heads in 10 tosses of the coin.

x Frequency of x Relative Frequency of x

x Frequency of x Relative Frequency of x

e Which value(s) of x occurred most frequently?

e If you tossed the coin 1,000 times, what values could x take on?
Which value(s) of z do you think would occur most frequently?

e What does the relative frequency column sum to?

Glossary

Random Variable (RV)
see Variable

Variable (Random Variable)
A characteristic of interest in a population being studied. Common
notation for variables are upper case Latin letters X, Y, Z,...; common
notation for a specific value from the domain (set of all possible values
of a variable) are lower case Latin letters x, y, z,.... For example, if X
is the number of children in a family, then x represents a specific
integer 0, 1, 2, 3, .... Variables in statistics differ from variables in
intermediate algebra in two following ways.

e The domain of the random variable (RV) is not necessarily a
numerical set; the domain may be expressed in words; for
example, if X = hair color then the domain is {black, blond, gray,
green, orange}.

e We can tell what specific value xz of the Random Variable X takes
only after performing the experiment.

Probability Distribution Function (PDF) for a Discrete Random Variable
This module introduces the Probability Distribution Function (PDF) and its
characteristics.

A discrete probability distribution function has two characteristics:

e Each probability is between 0 and 1, inclusive.
e The sum of the probabilities is 1.

Example:

A child psychologist is interested in the number of times a newborn baby's
crying wakes its mother after midnight. For a random sample of 50
mothers, the following information was obtained. Let _X = the number of
times a newborn wakes its mother after midnight. For this example, x = 0,
| Pare ac. ey

P(x) = probability that X takes on a value z.

x P(x)

0 ree) = =
1 P(ix=1) = _
2 x2) =
3 Puss) = =
4 Pix=4)— =

P(x=5) = 4

X takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because

1. Each P(x) is between 0 and 1, inclusive.
2. The sum of the probabilities is 1, that is,

Equation:

Zp lies eat a es
50 50 50 50 50 50,

Example:

Suppose Nancy has classes 3 days a week. She attends classes 3 days a
week 80% of the time, 2 days 15% of the time, 1 day 4% of the time, and

no days 1% of the time. Suppose one week is randomly selected.
Exercise:

Problem:

Let X = the number of days Nancy

Solution:

Let X = the number of days Nancy attends class per week.
Exercise:

Problem: X takes on what values?

Solution:

Oo. 2.-and:s

Exercise:

Problem:

Suppose one week is randomly chosen. Construct a probability
distribution table (called a PDF table) like the one in the previous
example. The table should have two columns labeled x and P(x).
What does the P(x) column sum to?

Solution:
x P(x)
0 0.01
1 0.04
2 0.15
3 0.80
Glossary

Probability Distribution Function (PDF)
A mathematical description of a discrete random variable (RV), given
either in the form of an equation (formula) , or in the form of a table
listing all the possible outcomes of an experiment and the probability
associated with each outcome.

Example:

A biased coin with probability 0.7 for a head (in one toss of the coin) is
tossed 5 times. We are interested in the number of heads (the RV X = the
number of heads). X is Binomial, so X ~ B(5,0.7) and P(X = x) =

5
( ) _77.3°-*or in the form of the table:
x

L P(x = 2)
0 0.0024
1 0.0284
2 0.1323
3 0.3087
4 0.3602

Mean or Expected Value and Standard Deviation

This module explores the Law of Large Numbers, the phenomenon where
an experiment performed many times will yield cumulative results closer
and closer to the theoretical mean over time.

The expected value is often referred to as the "long-term" average or
mean . This means that over the long term of doing an experiment over and
over, you would expect this average.

The mean of a random variable X is ps. If we do an experiment many times
(for instance, flip a fair coin, as Karl Pearson did, 24,000 times and let X =
the number of heads) and record the value of X each time, the average is
likely to get closer and closer to yz as we keep repeating the experiment.
This is known as the Law of Large Numbers.

Note:To find the expected value or long term average, jz, simply multiply
each value of the random variable by its probability and add the products.

A Step-by-Step Example

A men's soccer team plays soccer 0, 1, or 2 days a week. The probability
that they play O days is 0.2, the probability that they play 1 day is 0.5, and
the probability that they play 2 days is 0.3. Find the long-term average, jp,
or expected value of the days per week the men's soccer team plays soccer.

To do the problem, first let the random variable X = the number of days the
men's soccer team plays soccer per week. X takes on the values 0, 1, 2.
Construct a PDF table, adding a column xP(x). In this column, you will
multiply each z value by its probability.

x P(x) tP(x)

0 0.2 (0)(0.2) = 0
1 0.5 (1)(0.5) = 0.5
2 0.3 (2)(0.3) = 0.6

Expected Value TableThis table is called an expected value table. The table
helps you calculate the expected value or long-term average.

Add the last column to find the long term average or expected value:
(0)(0.2)+(1)(0.5)+(2)(0.3)= 0 + 0.5 + 0.6 = 1.1.

The expected value is 1.1. The men's soccer team would, on the average,
expect to play soccer 1.1 days per week. The number 1.1 is the long term
average or expected value if the men's soccer team plays soccer week after
week after week. We say H=1.1

Example:

Find the expected value for the example about the number of times a
newborn baby's crying wakes its mother after midnight. The expected
value is the expected number of times a newborn wakes its mother after
midnight.

2 P(X) xP(X

—_——

0 P(x=0) = (0)(s5) =0

1 P(x=1) = 5 (D(a) = 3
2 P(x=2) = #3 (2)(3>) = 3
3 PE) = a (3)(s5) = 3
4 Ria (4)(4)=-4
5 Poo, 6)(a)=4

You expect a newborn to wake its mother after midnight 2.1 times, on the
average.

Add the last column to find the expected value. jz = Expected Value =
105 __

9 = 2-1

Exercise:

Problem:

Go back and calculate the expected value for the number of days
Nancy attends classes a week. Construct the third column to do so.

Solution:

2.74 days a week.

Example:

Suppose you play a game of chance in which five numbers are chosen
from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A computer randomly selects five numbers
from 0 to 9 with replacement. You pay $2 to play and could profit
$100,000 if you match all 5 numbers in order (you get your $2 back plus

$100,000). Over the long term, what is your expected profit of playing the
game?

To do this problem, set up an expected value table for the amount of money
you can profit.

Let X = the amount of money you profit. The values of x are not 0, 1, 2, 3,
4, 5, 6, 7, 8, 9. Since you are interested in your profit (or loss), the values
of x are 100,000 dollars and -2 dollars.

To win, you must get all 5 numbers correct, in order. The probability of
choosing one correct number is ay because there are 10 numbers. You may
choose a number more than once. The probability of choosing all 5
numbers correctly and in order is:

Equation:

—_*__* __* ___*__*_ 1*19~° = 0.00001
10 10 10 10 10 ees

Therefore, the probability of winning is 0.00001 and the probability of
losing is
Equation:

1 — 0.00001 = 0.99999

The expected value table is as follows.

x P(x) xzP(x)
Loss 2 0.99999 (-2)(0.99999)=-1.99998
Profit 100,000 0.00001 (100000)(0.00001)=1

Add the last column. -1.99998 + 1 = -0.99998

Since —0.99998 is about —1, you would, on the average, expect to lose
approximately one dollar for each game you play. However, each time you
play, you either lose $2 or profit $100,000. The $1 is the average or
expected LOSS per game after playing this game over and over.

Example:
Suppose you play a game with a biased coin. You play each game by
tossing the coin once. P(heads) = 3 and P(tails) = 4. If you toss a

head, you pay $6. If you toss a tail, you win $10. If you play this game
many times, will you come out ahead?
Exercise:

Problem: Define a random variable X.

Solution:

X = amount of profit

Exercise:

Problem: Complete the following expected value table.

eo]

WIN 10

LOSE

Solution:

x P(x) xzP(x)
WIN 10 + 2
LOSE -6 2

Exercise:

Problem: What is the expected value, ? Do you come out ahead?

Solution:

Add the last column of the table. The expected value pz = =. You

lose, on average, about 67 cents each time you play the game so you
do not come out ahead.

Like data, probability distributions have standard deviations. To calculate
the standard deviation (@) of a probability distribution, find each deviation
from its expected value, square it, multiply it by its probability, add the
products, and take the square root . To understand how to do the calculation,
look at the table for the number of days per week a men's soccer team plays
soccer. To find the standard deviation, add the entries in the column labeled

(x — p)?- P(z) and take the square root.

z P(x) aP(x) (x -p)°P (x)

0 | o2 (0)(0.2) = 0 (0 — 1.1) (.2) ~ 0.242
1 05 (1)(0.5) = 0.5 Geta)? (.5) = 0.005
2 | 03 (2)(0.3) = 0.6 44) (.3) ~ 0.243

Add the last column in the table. 0.242 + 0.005 + 0.243 = 0.490. The
standard deviation is the square root of 0.49. 0 = / 0.49 — 0.7

Generally for probability distributions, we use a calculator or a computer to
calculate 4 and o to reduce roundoff error. For some probability
distributions, there are short-cut formulas that calculate jz and o.

Glossary

Expected Value
Expected arithmetic average when an experiment is repeated many
times. (Also called the mean). Notations: F(a), jw. For a discrete
random variable (RV) with probability distribution function P(a),the
definition can also be written in the form E(x) = uw = 5) xP(z).

Mean
A number that measures the central tendency. A common name for
mean is ‘average.’ The term 'mean' is a shortened form of ‘arithmetic
mean.’ By definition, the mean for a sample (denoted by 2) is
__ Sum of all values in the sample dth i lati
Lor Number of values in the sample ’ at € mean for a population
Sum of all values in the population
Number of values in the population °

(denoted by 4) is w =

Common Discrete Probability Distribution Functions

This module serves as a lead-in for several types of common discrete
probability distribution functions, including binomial, geometric,
hypergeometric, and Poisson.

Some of the more common discrete probability functions are binomial,
geometric, hypergeometric, and Poisson. Most elementary courses do not
cover the geometric, hypergeometric, and Poisson. Your instructor will let
you know if he or she wishes to cover these distributions.

A probability distribution function is a pattern. You try to fit a probability
problem into a pattern or distribution in order to perform the necessary
calculations. These distributions are tools to make solving probability
problems easier. Each distribution has its own special characteristics.
Learning the characteristics enables you to distinguish among the different
distributions.

The Binomial Random Variable

When you flip a coin, there are two possible outcomes: heads and tails.
Each outcome has a fixed probability, the same from trial to trial. In the
case of coins, heads and tails each have the same probability of 1/2. More
generally, there are situations in which the coin is biased, so that heads and
tails have different probabilities. In the present section, we consider
probability distributions for which there are just two possible outcomes
with fixed probability summing to one. These distributions are called are
called binomial distributions.

A Simple Example

The four possible outcomes that could occur if you flipped a coin twice are
listed in [link]. Note that the four outcomes are equally likely: each has
probability 1/4. To see this, note that the tosses of the coin are independent
(neither affects the other). Hence, the probability of a head on Flip 1 anda
head on Flip 2 is the product of P{H] and P[H], which is 1/2 x 1/2 = 1/4.
The same calculation applies to the probability of a head on Flip one and a
tail on Flip 2. Each is 1/2 x 1/2 = 1/4.

Outcome First Flip Second Flip
uf Heads Heads

2 Heads Tails

3 Tails Heads

4 Tails Tails

Four Possible Outcomes

The four possible outcomes can be classifid in terms of the number of heads
that come up. The number could be two (Outcome 1), one (Outcomes 2 and
3) or 0 (Outcome 4). The probabilities of these possibilities are shown in
[link] and in [link]. Since two of the outcomes represent the case in which
just one head appears in the two tosses, the probability of this event is equal
to1/4+ 1/4 = 1/2. [link] summarizes the situation.

Number of Heads Probability
0 1/4
1 1/2
2 1/4

Probabilities of Getting 0,1, or 2 heads.

0.55
0.50
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00

Probability

1
Number of Heads

Probabilities of 0, 1, and 2 heads.

[link] is a discrete probability distribution: It shows the probability for each
of the values on the X-axis. Defining a head as a "success," [link] shows the
probability of 0, 1, and 2 successes for two trials (flips) for an event that has
a probability of 0.5 of being a success on each trial. This makes [link] an
example of a binomial distribution.

The Formula for Binomial Probabilities

The binomial distribution consists of the probabilities of each of the
possible numbers of successes on 7 trials for independent events that each
have a probability of p of occurring. For the coin flip example, n = 2 and
p=(0.5). The formula for the binomial distribution is shown below:

Phe) = Sra)" *

where P|x] is the probability of x successes out of 7 trials, n is the number
of trials, and p is the probability of success on a given trial. Applying this to
the coin flip example,

2! i.

P[0] = ——0.5°(1 — 0.5)?" ° = =1 x .25 = 0.25
0! (2 —0)! 2
2! 42

P{1] = ————_0.51(1 — 0.5)? 1 = —.5 x .5 = 0.50
He—1)) 1
2! 2

P[2] = ———___0.5?(1 — 0.5)?-? = —.25 x 1 = 0.25
HOH) 2

If you flip a coin twice, what is the probability of getting one or more
heads? Since the probability of getting exactly one head is 0.50 and the
probability of getting exactly two heads is 0.25, the probability of getting
one or more heads is 0.50 + 0.25 = 0.75.

Now suppose that the coin is biased; let's say the probability of heads is
only 0.4. What is the probability of getting heads at least once in two

tosses? We could substitute p=0.4 with x=1 and with x=2 into our general
formula above; adding the results would obtain the answer 0.64.

Cumulative Probabilities

We toss a coin 12 times. What is the probability that we get from 0 to 3
heads? The answer is found by computing the probability of exactly 0
heads, exactly 1 head, exactly 2 heads, and exactly 3 heads. The probability
of getting from 0 to 3 heads is then the sum of these probabilities. The
probabilities are: 0.0002, 0.0029, 0.0161, and 0.0537. The sum of the
probabilities is 0.073. The calculation of cumulative binomial probabilities
can be quite tedious. Therefore we have provided a binomial calculator to
make it easy to calculate these probabilities.

Note: Click here for the binomial calculator.

Mean and Standard Deviation of Binomial Distributions

Consider a coin-tossing experiment in which you tossed a coin 12 times and
recorded the number of heads. If you performed this experiment over and
over again, what would the mean number of heads be? On average, you
would expect half the coin tosses to come up heads. Therefore the mean
number of heads would be 6. In general, the mean of a binomial distribution
with parameters m (the number of trials) and p (the probability of success
for each trial) is:

p= np

where yz is the mean of the binomial distribution. The variance of the
binomial distribution is:

o* = np(1—p)

where o” is the variance of the binomial distribution.

Let's return to the coin tossing experiment. The coin was tossed 12 times so
nm = 12. A coin has a probability of 0.5 of coming up heads. Therefore,

p = 0.5. The mean and standard deviation can therefore be computed as
follows:

p=np =12x0.5=6
o? = np(1—p) = 12 x 0.5 x (1.0 — 0.5) = 3.0

Naturally, the standard deviation (c) is the square root of the variance (0).

Binomial Calculator

Applet failed to run. No Java plug-in was found.

Binomial Experiments
This module describes the characteristics of a binomial experiment and the
binomial probability distribution function.

The characteristics of a binomial experiment are:

1.

Z.

There are a fixed number of trials. Think of trials as repetitions of an
experiment. The letter m denotes the number of trials.

There are only 2 possible outcomes, called "success" and, "failure" for
each trial. The letter p denotes the probability of a success on one trial
and q denotes the probability of a failure on one trial. p + q = 1.

. The 7 trials are independent and are repeated using identical

conditions. Because the n trials are independent, the outcome of one
trial does not help in predicting the outcome of another trial. Another
way of saying this is that for each individual trial, the probability, p, of
a success and probability, g, of a failure remain the same. For example,
randomly guessing at a true - false statistics question has only two
outcomes. If a success is guessing correctly, then a failure is guessing
incorrectly. Suppose Joe always guesses correctly on any statistics true
- false question with probability p = 0.6. Then, q = 0.4 .This means
that for every true - false statistics question Joe answers, his
probability of success (p = 0.6) and his probability of failure (q = 0.4
) remain the same.

The outcomes of a binomial experiment fit a binomial probability
distribution. The random variable X = the number of successes obtained
in the n independent trials.

The mean, jz, and variance, o”, for the binomial probability distribution is
y. = np and o? = npq. The standard deviation, a, is then o = ,/npq.

Any experiment that has characteristics 2 and 3 and where n = 1 is calleda
Bernoulli Trial (named after Jacob Bernoulli who, in the late 1600s,
studied them extensively). A binomial experiment takes place when the
number of successes is counted in one or more Bernoulli Trials.

Example:

At ABC College, the withdrawal rate from an elementary physics course is
30% for any given term. This implies that, for any given term, 70% of the
students stay in the class for the entire term. A "success" could be defined
as an individual who withdrew. The random variable is X = the number of
students who withdraw from the randomly selected elementary physics
class.

Example:

Suppose you play a game that you can only either win or lose. The
probability that you win any game is 55% and the probability that you lose
is 45%. Each game you play is independent. If you play the game 20 times,
what is the probability that you win 15 of the 20 games? Here, if you
define X = the number of wins, then X takes on the values 0, 1, 2, 3, ...,
20. The probability of a success is p = 0.55. The probability of a failure is
q = 0.45. The number of trials ism = 20. The probability question can be
stated mathematically as P(a = 15).

Example:

A fair coin is flipped 15 times. Each flip is independent. What is the
probability of getting more than 10 heads? Let X = the number of heads in
15 flips of the fair coin. X takes on the values 0, 1, 2, 3, ..., 15. Since the
coin is fair, p = 0.5 and q = 0.5. The number of trials is n = 15. The
probability question can be stated mathematically as P(a > 10).

Example:

Approximately 70% of statistics students do their homework in time for it
to be collected and graded. Each student does homework independently. In
a Statistics class of 50 students, what is the probability that at least 40 will
do their homework on time? Students are selected randomly.

Exercise:

Problem:

This is a binomial problem because there is only a success or a
, there are a definite number of trials, and the probability
of a success is 0.70 for each trial.

Solution:

failure
Exercise:

Problem:

If we are interested in the number of students who do their homework,
then how do we define X?

Solution:

X =the number of statistics students who do their homework on time
Exercise:

Problem: What values does zx take on?

Solution:
Ole 0)
Exercise:

Problem: What is a "failure", in words?

Solution:

Failure is a student who does not do his or her homework on time.

The probability of a success is p = 0.70. The number of trial is n = 50.
Exercise:

Problem: If p + gq = 1, then what is q?
Solution:

gq = 0.30
Exercise:
Problem:

The words "at least" translate as what kind of inequality for the
probability question P(a____40).

Solution:

greater than or equal to (=)

The probability question is P(a > 40).

Notation for the Binomial: B = Binomial Probability
Distribution Function

X ~ B(n,p)

Read this as ".X is arandom variable with a binomial distribution." The
parameters are n and p. n = number of trials p = probability of a success on
each trial

Example:

It has been stated that about 41% of adult workers have a high school
diploma but do not pursue any further education. If 20 adult workers are
randomly selected, find the probability that at most 12 of them have a high
school diploma but do not pursue any further education. How many adult

workers do you expect to have a high school diploma but do not pursue
any further education?

Let X = the number of workers who have a high school diploma but do not
pursue any further education.

X takes on the values 0, 1, 2, ..., 20 where n = 20 and p = 0.41. q=1-
0.41 = 0.59. X ~ B(20, 0.41)

Find P(x < 12). P(a < 12) = 0.9738. (calculator or computer)

Using the TI-83+ or the TI-84 calculators, the calculations are as follows.
Go into 2nd DISTR. The syntax for the instructions are

To calculate (x = value): binompdf(n, p, number) If "number" is left out,
the result is the binomial probability table.

To calculate P(x < value): binomcdf(n, p, number) If "number" is left
out, the result is the cumulative binomial probability table.

For this problem: After you are in 2nd DISTR, arrow down to
binomcdf. Press ENTER. Enter 20,.41,12). The result is

(12 O88:

Note:If you want to find P(a = 12), use the pdf (binompdf). If you want
to find P(x>12), use 1 - binomcdf(20,.41,12).

The probability at most 12 workers have a high school diploma but do not
pursue any further education is 0.9738
The graph of x ~ B(20, 0.41) is:

0123 4 Sires» 20

The y-axis contains the probability of 2, where X = the number of workers
who have only a high school diploma.

The number of adult workers that you expect to have a high school
diploma but not pursue any further education is the mean,

fe — np —"(20)( OAT) 62;

The formula for the variance is o“ = npq. The standard deviation is

o = ./upq. o = »/(20)(0.41)(0.59) = 2.20.

2

Example:

The following example illustrates a problem that is not binomial. It
violates the condition of independence. ABC College has a student
advisory committee made up of 10 staff members and 6 students. The
committee wishes to choose a chairperson and a recorder. What is the
probability that the chairperson and recorder are both students? All names
of the committee are put into a box and two names are drawn without
replacement. The first name drawn determines the chairperson and the
second name the recorder. There are two trials. However, the trials are not
independent because the outcome of the first trial affects the outcome of
the second trial. The probability of a student on the first draw is +. The

probability of a student on the second draw is — when the first draw
produces a student. The probability is a when the first draw produces a

staff member. The probability of drawing a student's name changes for
each of the trials and, therefore, violates the condition of independence.

Glossary

Bernoulli Trials
An experiment with the following characteristics:

e There are only 2 possible outcomes called “success” and “failure”
for each trial.

e The probability p of a success is the same for any trial (so the
probability g = 1 — p of a failure is the same for any trial).

Binomial Distribution
A discrete random variable (RV) which arises from Bernoulli trials.
There are a fixed number, n, of independent trials. “Independent”
means that the result of any trial (for example, trial 1) does not affect
the results of the following trials, and all trials are conducted under the
same conditions. Under these circumstances the binomial RV X is
defined as the number of successes in n trials. The notation is: X~
B(n, p). The mean is js = np and the standard deviation is o = ,/npq
. The probability of exactly z successes in 7 trials is

P(X = 2) = (2 )p*q"*.

Summary of the Binomial Formulas

This module provides a review of the binomial, geometric, hypergeometric,
and Poisson probability distribution functions and their properties.
Formula

Binomial

X~B(n,p)

X =the number of successes in n independent trials
m = the number of independent trials

X takes on the values z = 0,1, 2, 3, ...,n

p = the probability of a success for any trial

q = the probability of a failure for any trial
Parga. GaLop

The mean is 2 = np. The standard deviation is 0 = ,/npq.

Practice 1: Discrete Distribution

This module provides students an opportunity to practice applying concepts
related to discrete distributions. This practice exercise asks students to
calculate several values based on the data provided.

Student Learning Outcomes

e The student will analyze the properties of a discrete distribution.

Given:

A ballet instructor is interested in knowing what percent of each year's class
will continue on to the next, so that she can plan what classes to offer. Over
the years, she has established the following probability distribution.

e Let X =the number of years a student will study ballet with the

teacher.
e Let P(x) = the probability that a student will study ballet x years.

Organize the Data

Complete the table below using the data provided.

x P(x) x*P(x)
1 0.10
2 0.05

3 0.10

x P(x) x*P(x)

4

5 0.30

6 0.20

Zz 0.10
Exercise:

Problem: In words, define the Random Variable X.

Exercise:

Problem: P(x = 4) =

Exercise:

Problem: P(x < 4) =
Exercise:

Problem:
On average, how many years would you expect a child to study ballet
with this teacher?

Discussion Question

Exercise:

Problem: What does the column "P(x)" sum to and why?

Exercise:

Problem: What does the column "«*P(x)" sum to and why?

Practice 2: Binomial Distribution

This module provides a practice of Binomial Distribution as a part of
Collaborative Statistics collection (col10522) by Barbara Illowsky and
Susan Dean.

Student Learning Outcomes

e The student will construct the Binomial Distribution.

Given

The Higher Education Research Institute at UCLA collected data from
203,967 incoming first-time, full-time freshmen from 270 four-year
colleges and universities in the U.S. 71.3% of those students replied that,
yes, they believe that same-sex couples should have the right to legal
marital status. (Source:
http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/TheAmericanFres
hman2011.pdf). )

Suppose that you randomly pick 8 first-time, full-time freshmen from the

survey. You are interested in the number that believes that same sex-couples
should have the right to legal marital status

Interpret the Data

Exercise:

Problem: In words, define the random Variable X.

Solution:

X= the number that reply “yes”

Exercise:

Problem: X~

Solution:
B(8,0.713)

Exercise:

Problem: What values does the random variable X take on?

Solution:
0,1,2.5:4:5;6;7,8

Exercise:

Problem: Construct the probability distribution function (PDF).

Exercise:

Problem:

On average (u), how many would you expect to answer yes?

Solution:

D7

Exercise:

Problem: What is the standard deviation (c) ?

Solution:

1.28
Exercise:

Problem:

What is the probability that at most 5 of the freshmen reply “yes”?

Solution:

0.4151
Exercise:

Problem:
What is the probability that at least 2 of the freshmen reply “yes”?
Solution:

0.9990
Exercise:

Problem:

Construct a histogram or plot a line graph. Label the horizontal and
vertical axes with words. Include numerical scaling.

Homework

Discrete Random Variables: Homework is part of the collection col10555
written by Barbara Illowsky and Susan Dean Homework and provides a
number of homework exercises related to Discrete Random Variables
(binomial, geometric, hypergeometric and Poisson) with contributions from
Roberta Bloom.

Exercise:

Problem: 1. Complete the PDF and answer the questions.

£ P(X = 2) POX a)
0 0.3

1 0.2

Z

3 0.4

e aFind the probability that z = 2.
e b Find the expected value.

Solution:

e a0.1
e b1.6

Exercise:

Problem:

Suppose that you are offered the following “deal.” You roll a die. If
you roll a 6, you win $10. If you roll a 4 or 5, you win $5. If you roll a
1, 2, or 3, you pay $6.

e¢ aWhat are you ultimately interested in here (the value of the roll
or the money you win)?

e bln words, define the Random Variable X.

e cList the values that X may take on.

e dConstruct a PDF.

e eOver the long run of playing this game, what are your expected
average winnings per game?

e fBased on numerical values, should you take the deal? Explain
your decision in complete sentences.

Exercise:

Problem:

A venture capitalist, willing to invest $1,000,000, has three
investments to choose from. The first investment, a software company,
has a 10% chance of returning $5,000,000 profit, a 30% chance of
returning $1,000,000 profit, and a 60% chance of losing the million
dollars. The second company, a hardware company, has a 20% chance
of returning $3,000,000 profit, a 40% chance of returning $1,000,000
profit, and a 40% chance of losing the million dollars. The third
company, a biotech firm, has a 10% chance of returning $6,000,000
profit, a 70% of no profit or loss, and a 20% chance of losing the
million dollars.

aConstruct a PDF for each investment.

bFind the expected value for each investment.

cWhich is the safest investment? Why do you think so?
dWhich is the riskiest investment? Why do you think so?
eWhich investment has the highest expected return, on average?

Solution:

¢ b$200,000;$600,000;$400,000
e cthird investment

e dfirst investment

e esecond investment

Exercise:

Problem:

A theater group holds a fund-raiser. It sells 100 raffle tickets for $5
apiece. Suppose you purchase 4 tickets. The prize is 2 passes to a
Broadway show, worth a total of $150.

e aWhat are you interested in here?

e bln words, define the Random Variable X.

e cList the values that X may take on.

e dConstruct a PDF.

e elf this fund-raiser is repeated often and you always purchase 4
tickets, what would be your expected average winnings per raffle?

Exercise:

Problem:

Suppose that 20,000 married adults in the United States were randomly
surveyed as to the number of children they have. The results are
compiled and are used as theoretical probabilities. Let X = the number
of children

0 0.10
1 0.20
2 0.30
3

4 0.10
5 0.05
6 (or more) 0.05

aFind the probability that a married adult has 3 children.

bIn words, what does the expected value in this example
represent?

c Find the expected value.

d Is it more likely that a married adult will have 2 — 3 children or
4 —6 children? How do you know?

Solution:

a0.2
€2:35
d2-3 children

Exercise:

Problem:

Suppose that the PDF for the number of years it takes to earn a
Bachelor of Science (B.S.) degree is given below.

3 0.05
4 0.40
is) 0.30
6 0.15
7 0.10

e aln words, define the Random Variable X.

e b What does it mean that the values 0, 1, and 2 are not included
for x in the PDF?

e cOn average, how many years do you expect it to take for an
individual to earn a B.S.?

For each problem:

e aln words, define the Random Variable X.
¢ bList the values that X may take on.
e cGive the distribution of X. X~

Then, answer the questions specific to each individual problem.
Exercise:
Problem:

Six different colored dice are rolled. Of interest is the number of dice
that show a “1.”

e dOn average, how many dice would you expect to show a “1”?
e eFind the probability that all six dice show a “1.”

e fls it more likely that 3 or that 4 dice will show a “1”? Use
numbers to justify your answer numerically.

Solution:

e aX =the number of dice that show a 1
¢ b0,1,2,3,4,5,6

¢ c X~B(6, =)

edi

e e 0.00002

e f3 dice

Exercise:

Problem:

More than 96 percent of the very largest colleges and universities
(more than 15,000 total enrollments) have some online offerings.
Suppose you randomly pick 13 such institutions. We are interested in
the number that offer distance learning courses. (Source:
http://en.wikipedia.org/wiki/Distance_education)

e dOn average, how many schools would you expect to offer such
courses?

e eFind the probability that at most 6 offer such courses.

e fls it more likely that 0 or that 13 will offer such courses? Use
numbers to justify your answer numerically and answer in a
complete sentence.

Exercise:

Problem:

A school newspaper reporter decides to randomly survey 12 students
to see if they will attend Tet (Vietnamese New Year) festivities this
year. Based on past years, she knows that 18% of students attend Tet
festivities. We are interested in the number of students who will attend
the festivities.

e dHow many of the 12 students do we expect to attend the
festivities?

e eFind the probability that at most 4 students will attend.

e fFind the probability that more than 2 students will attend.

Solution:

e a X =the number of students that will attend Tet.
s- DO; 12. 3..4.°5.67 7,8, 9. 10: 1112

e c X~B(12,0.18)

e d2.16

e e0.9511

e £0.3702

Exercise:
Problem:

Suppose that about 85% of graduating students attend their graduation.
A group of 22 graduating students is randomly chosen.

e dHow many are expected to attend their graduation?

e eFind the probability that 17 or 18 attend.

e fBased on numerical values, would you be surprised if all 22
attended graduation? Justify your answer numerically.

Exercise:

Problem:

At The Fencing Center, 60% of the fencers use the foil as their main
weapon. We randomly survey 25 fencers at The Fencing Center. We
are interested in the numbers that do not use the foil as their main
weapon.

e dHow many are expected to not use the foil as their main
weapon?

e eFind the probability that six do not use the foil as their main
weapon.

e fBased on numerical values, would you be surprised if all 25 did
not use foil as their main weapon? Justify your answer
numerically.

Solution:

e aX =the number of fencers that do not use foil as their main
weapon

e bO, 1, 2, 3,... 25

¢ c X~B(25,0.40)

e di0

e e0.0442

e fYes

Exercise:

Problem:

Approximately 8% of students at a local high school participate in
after-school sports all four years of high school. A group of 60 seniors
is randomly chosen. Of interest is the number that participated in after-
school sports all four years of high school.

e dHow many seniors are expected to have participated in after-
school sports all four years of high school?

e eBased on numerical values, would you be surprised if none of
the seniors participated in after-school sports all four years of
high school? Justify your answer numerically.

e fBased upon numerical values, is it more likely that 4 or that 5 of
the seniors participated in after-school sports all four years of
high school? Justify your answer numerically.

Exercise:

Problem:

The chance of having an extra fortune in a fortune cookie is about 3%.
Given a bag of 144 fortune cookies, we are interested in the number of
cookies with an extra fortune. Two distributions may be used to solve
this problem. Use one distribution to solve the problem.

e dHow many cookies do we expect to have an extra fortune?

e eFind the probability that none of the cookies have an extra
fortune.

e fFind the probability that more than 3 have an extra fortune.

e gAs n increases, what happens involving the probabilities using
the two distributions? Explain in complete sentences.

Solution:

e aX =the number of fortune cookies that have an extra fortune
e bO, 1, 2, 3,... 144

¢ c X~B(144, 0.03) or P(4.32)

e d4.32

e e 0.0124 or 0.0133

e f 0.6300 or 0.6264

Exercise:

Problem:

There are two games played for Chinese New Year and Vietnamese
New Year. They are almost identical. In the Chinese version, fair dice
with numbers 1, 2, 3, 4, 5, and 6 are used, along with a board with
those numbers. In the Vietnamese version, fair dice with pictures of a
gourd, fish, rooster, crab, crayfish, and deer are used. The board has
those six objects on it, also. We will play with bets being $1. The
player places a bet on a number or object. The “house” rolls three dice.
If none of the dice show the number or object that was bet, the house
keeps the $1 bet. If one of the dice shows the number or object bet
(and the other two do not show it), the player gets back his $1 bet, plus
$1 profit. If two of the dice show the number or object bet (and the
third die does not show it), the player gets back his $1 bet, plus $2
profit. If all three dice show the number or object bet, the player gets
back his $1 bet, plus $3 profit.

Let X = number of matches and Y= profit per game.

e dList the values that Y may take on. Then, construct one PDF
table that includes both X & Y and their probabilities.

e eCalculate the average expected matches over the long run of
playing this game for the player.

e fCalculate the average expected earnings over the long run of
playing this game for the player.

e gDetermine who has the advantage, the player or the house.

Exercise:

Problem:

According to the South Carolina Department of Mental Health web
site, for every 200 U.S. women, the average number who suffer from
anorexia is one (http:/Awww.state.sc.us/dmh/anorexia/statistics.htm).
Out of a randomly chosen group of 600 U.S. women:

¢ dHow many are expected to suffer from anorexia?
e eFind the probability that no one suffers from anorexia.

e fFind the probability that more than four suffer from anorexia.

Solution:

e a X =the number of women that suffer from anorexia
e bO, 1, 2, 3,... 600 (can leave off 600)

e c X~P(3)

e d3

e e0.0498

e £0.1847

Exercise:

Problem:

The average number of children a Japanese woman has in her lifetime
is 1.37. Suppose that one Japanese woman is randomly chosen.
(http://www.mhlw.go.jp/english/policy/children/children-
childrearing/index.html MHLW’s Pamphlet)

e dFind the probability that she has no children.

e eFind the probability that she has fewer children than the
Japanese average.

e fFind the probability that she has more children than the Japanese
average.

Exercise:

Problem:

The average number of children a Spanish woman has in her lifetime
is 1.47. Suppose that one Spanish woman is randomly chosen.
(http://www.typicallyspanish.com/news/publish/article 4897.shtml).

e dFind the probability that she has no children.
e eFind the probability that she has fewer children than the Spanish
average.

e fFind the probability that she has more children than the Spanish
average .

Solution:

e a X =the number of children for a Spanish woman
¢ bo, 1, 2, 3....

e c X~P(1.47)

e d0.2299

° e0.5679

e 0.4321

Exercise:

Problem:

Fertile (female) cats produce an average of 3 litters per year. (Source:
The Humane Society of the United States). Suppose that one fertile,
female cat is randomly chosen. In one year, find the probability she
produces:

e dNo litters.
e eAt least 2 litters.
e fExactly 3 litters.

Exercise:

Problem:

A consumer looking to buy a used red Miata car will call dealerships
until she finds a dealership that carries the car. She estimates the
probability that any independent dealership will have the car will be
28%. We are interested in the number of dealerships she must call.

e dOn average, how many dealerships would we expect her to have
to call until she finds one that has the car?
e eFind the probability that she must call at most 4 dealerships.

e fFind the probability that she must call 3 or 4 dealerships.

Solution:

e a X =the number of dealers she calls until she finds one with a
used red Miata

e bi, 2, 3...

e c X~G(0.28)

e d3.57

e e0.7313

e £0.2497

Exercise:

Problem:

Suppose that the probability that an adult in America will watch the
Super Bowl is 40%. Each person is considered independent. We are
interested in the number of adults in America we must survey until we
find one who will watch the Super Bowl.

e dHow many adults in America do you expect to survey until you
find one who will watch the Super Bowl?

e eFind the probability that you must ask 7 people.

e fFind the probability that you must ask 3 or 4 people.

Exercise:

Problem:

A group of Martial Arts students is planning on participating in an
upcoming demonstration. 6 are students of Tae Kwon Do; 7 are
students of Shotokan Karate. Suppose that 8 students are randomly
picked to be in the first demonstration. We are interested in the number
of Shotokan Karate students in that first demonstration. Hint: Use the
Hypergeometric distribution. Look in the Formulas section of 4:
Discrete Distributions and in the Appendix Formulas.

e dHow many Shotokan Karate students do we expect to be in that
first demonstration?

e eFind the probability that 4 students of Shotokan Karate are
picked for the first demonstration.

e fSuppose that we are interested in the Tae Kwan Do students that
are picked for the first demonstration. Find the probability that all
6 students of Tae Kwan Do are picked for the first demonstration.

Solution:

e d4.31
e e0.4079
¢ £0.0163

Exercise:

Problem:

The chance of a IRS audit for a tax return with over $25,000 in income
is about 2% per year. We are interested in the expected number of
audits a person with that income has in a 20 year period. Assume each
year is independent.

e dHow many audits are expected in a 20 year period?
e eFind the probability that a person is not audited at all.
e fFind the probability that a person is audited more than twice.

Exercise:

Problem:

Refer to the previous problem. Suppose that 100 people with tax
returns over $25,000 are randomly picked. We are interested in the
number of people audited in 1 year. One way to solve this problem is
by using the Binomial Distribution. Since n is large and p is small,
another discrete distribution could be used to solve the following
problems. Solve the following questions (d-f) using that distribution.

e dHow many are expected to be audited?
e eFind the probability that no one was audited.
e fFind the probability that more than 2 were audited.

Solution:

e d2
e e0.1353
¢ £0.3233

Exercise:

Problem:

Suppose that a technology task force is being formed to study
technology awareness among instructors. Assume that 10 people will
be randomly chosen to be on the committee from a group of 28
volunteers, 20 who are technically proficient and 8 who are not. We
are interested in the number on the committee who are not technically
proficient.

e dHow many instructors do you expect on the committee who are
not technically proficient?

e eFind the probability that at least 5 on the committee are not
technically proficient.

e fFind the probability that at most 3 on the committee are not
technically proficient.

Exercise:
Problem:

Refer back to Exercise 4.15.12. Solve this problem again, using a
different, though still acceptable, distribution.

Solution:

e a X =the number of seniors that participated in after-school
sports all 4 years of high school

e b0, 1, 2, 3,... 60

e c X~P(4.8)

e d4.8

e eYes

e £4

Exercise:

Problem:

Suppose that 9 Massachusetts athletes are scheduled to appear at a
charity benefit. The 9 are randomly chosen from 8 volunteers from the
Boston Celtics and 4 volunteers from the New England Patriots. We
are interested in the number of Patriots picked.

e dis it more likely that there will be 2 Patriots or 3 Patriots picked?

Exercise:

Problem:

On average, Pierre, an amateur chef, drops 3 pieces of egg shell into
every 2 batters of cake he makes. Suppose that you buy one of his
cakes.

e dOn average, how many pieces of egg shell do you expect to be
in the cake?

e eWhat is the probability that there will not be any pieces of egg
shell in the cake?

e fLet’s say that you buy one of Pierre’s cakes each week for 6
weeks. What is the probability that there will not be any egg shell
in any of the cakes?

e gBased upon the average given for Pierre, is it possible for there
to be 7 pieces of shell in the cake? Why?

Solution:

a X = the number of shell pieces in one cake
bO, 1, 2, 3....

c X~P(1.5)

di.5

e0.2231

£0.0001

gYes

Exercise:

Problem:

It has been estimated that only about 30% of California residents have
adequate earthquake supplies. Suppose we are interested in the number
of California residents we must survey until we find a resident who
does not have adequate earthquake supplies.

dWhat is the probability that we must survey just 1 or 2 residents
until we find a California resident who does not have adequate
earthquake supplies?

eWhat is the probability that we must survey at least 3 California
residents until we find a California resident who does not have
adequate earthquake supplies?

fHow many California residents do you expect to need to survey
until you find a California resident who does not have adequate
earthquake supplies?

gHow many California residents do you expect to need to survey
until you find a California resident who does have adequate
earthquake supplies?

Exercise:

Problem:

Refer to the above problem. Suppose you randomly survey 11
California residents. We are interested in the number who have
adequate earthquake supplies.

e d What is the probability that at least 8 have adequate earthquake
supplies?

e els it more likely that none or that all of the residents surveyed
will have adequate earthquake supplies? Why?

e fHow many residents do you expect will have adequate
earthquake supplies?

Solution:

e d0.0043
e enone
* 433

The next 2 questions refer to the following: In one of its Spring catalogs,
L.L. Bean® advertised footwear on 29 of its 192 catalog pages.
Exercise:

Problem:

Suppose we randomly survey 20 pages. We are interested in the
number of pages that advertise footwear. Each page may be picked at
most once.

e dHow many pages do you expect to advertise footwear on them?

¢ els it probable that all 20 will advertise footwear on them? Why
or why not?

e fWhat is the probability that less than 10 will advertise footwear
on them?

Exercise:

Problem:

Suppose we randomly survey 20 pages. We are interested in the
number of pages that advertise footwear. This time, each page may be
picked more than once.

dHow many pages do you expect to advertise footwear on them?
els it probable that all 20 will advertise footwear on them? Why
or why not?

f What is the probability that less than 10 will advertise footwear
on them?

gReminder: A page may be picked more than once. We are
interested in the number of pages that we must randomly survey
until we find one that has footwear advertised on it. Define the
random variable X and give its distribution.

hWhat is the probability that you only need to survey at most 3
pages in order to find one that advertises footwear on it?

iHow many pages do you expect to need to survey in order to find
one that advertises footwear?

Solution:

d3.02

eNo

£0.9997
h0.3881
16.6207 pages

Exercise:

Problem:

Suppose that you roll a fair die until each face has appeared at least
once. It does not matter in what order the numbers appear. Find the
expected number of rolls you must make until each face has appeared
at least once.

Try these multiple choice problems.

For the next three problems: The probability that the San Jose Sharks will
win any given game is 0.3694 based on a 13 year win history of 382 wins
out of 1034 games played (as of a certain date). An upcoming monthly
schedule contains 12 games.

Let X = the number of games won in that upcoming month.

Exercise:

Problem: The expected number of wins for that upcoming month is:

¢ A1.67

° Bl2

382
° C iB

e D443

Solution:

D: 4.43
Exercise:

Problem:

What is the probability that the San Jose Sharks win 6 games in that
upcoming month?

e A0.1476
e B0.2336
e C0.7664
e DO0.8903

Solution:

A: 0.1476

Exercise:

Problem:

What is the probability that the San Jose Sharks win at least 5 games in
that upcoming month

e A0.3694
e B0.5266
e C0.4734
e DO.2305

Solution:

C: 0.4734

For the next two questions: The average number of times per week that
Mrs. Plum’s cats wake her up at night because they want to play is 10. We
are interested in the number of times her cats wake her up each week.
Exercise:

Problem: In words, the random variable X =

e A The number of times Mrs. Plum’s cats wake her up each week
¢ B The number of times Mrs. Plum’s cats wake her up each hour

e C The number of times Mrs. Plum’s cats wake her up each night
¢ D The number of times Mrs. Plum’s cats wake her up

Solution:

A: The number of times Mrs. Plum's cats wake her up each week
Exercise:
Problem:

Find the probability that her cats will wake her up no more than 5
times next week.

e A0.5000
e B0.9329
e €0.0378
e DO0.0671

Solution:

D: 0.0671
Exercise:

Problem:

People visiting video rental stores often rent more than one DVD at a
time. The probability distribution for DVD rentals per customer at
Video To Go is given below. There is 5 video limit per customer at this
store, so nobody ever rents more than 5 DVDs.

x 0 1 2 3 4 fs)

P(X=x) 0.03 O50 0.24 #? 0.07 & 0.04

e A Describe the random variable X in words.

e B Find the probability that a customer rents three DVDs.

e C Find the probability that a customer rents at least 4 DVDs.
e D Find the probability that a customer rents at most 2 DVDs.

Another shop, Entertainment Headquarters, rents DVDs and
videogames. The probability distribution for DVD rentals per customer
at this shop is given below. They also have a5 DVD limit per
customer.

x 0 1 2 3 4 fs)

P(X=x) 035 0.25 0.20 010 0.05 0.05

e E At which store is the expected number of DVDs rented per
customer higher?

e F If Video to Go estimates that they will have 300 customers next
week, how many DVDs do they expect to rent next week?
Answer in sentence form.

¢ GIf Video to Go expects 300 customers next week and
Entertainment HQ projects that they will have 420 customers, for
which store is the expected number of DVD rentals for next week
higher? Explain.

e H Which of the two video stores experiences more variation in
the number of DVD rentals per customer? How do you know
that?

Solution:

Partial Answer:

A: X = the number of DVDs a Video to Go customer rents
B: 0.12

C: 0.11

D: 0.77

Exercise:
Problem:
A game involves selecting a card from a deck of cards and tossing a
coin. The deck has 52 cards and 12 cards are "face cards" (Jack,

Queen, or King) The coin is a fair coin and is equally likely to land on
Heads or Tails

e If the card is a face card and the coin lands on Heads, you win $6
e If the card is a face card and the coin lands on Tails, you win $2

e If the card is not a face card, you lose $2, no matter what the coin
shows.

e A Find the expected value for this game (expected net gain or
loss).

¢ B Explain what your calculations indicate about your long-term
average profits and losses on this game.

e¢ C Should you play this game to win money?

Solution:

The variable of interest is X = net gain or loss, in dollars

The face cards J, Q, K (Jack, Queen, King). There are(3)(4) = 12 face
cards and 52 — 12 = 40 cards that are not face cards.

We first need to construct the probability distribution for X. We use the
card and coin events to determine the probability for each outcome,
but we use the monetary value of X to determine the expected value.

$X net gain or

Card Event gee P(X)
(12/52)(1/2) =
Face Card and Heads 6 6/52
(12/52)(1/2) =
Face Card and Tails 2 6/52
(Not Face Card) and (H -) (40/52)(1) =

or T) 40/52

e Expected value = (6)(6/52) + (2)(6/52) + (—2) (40/52) = —32/52

e Expected value = —$0.62, rounded to the nearest cent

e If you play this game repeatedly, over a long number of games,
you would expect to lost 62 cents per game, on average.

e You should not play this game to win money because the
expected value indicates an expected average loss.

Exercise:
Problem:
You buy a lottery ticket to a lottery that costs $10 per ticket. There are
only 100 tickets available be sold in this lottery. In this lottery there is

one $500 prize, 2 $100 prizes and 4 $25 prizes. Find your expected
gain or loss.

Solution:

Start by writing the probability distribution. X is net gain or loss =
prize (if any) less $10 cost of ticket

X = $ net gain or loss P(X)

$500-—$10=$490 1/100
$100—$10=$90 2/100
$25-$10=$15 4/100

$0-$10=$-10 93/100)

Expected Value = (490)(1/100) + (90)(2/100) + (15)(4/100) + (—10)
(93/100) = —$2. There is an expected loss of $2 per ticket, on average.

Exercise:

Problem:

A student takes a 10 question true-false quiz, but did not study and
randomly guesses each answer. Find the probability that the student
passes the quiz with a grade of at least 70% of the questions correct.

Solution:

e X =number of questions answered correctly

e X~B(10, 0.5)

e We are interested in AT LEAST 70% of 10 questions correct.
70% of 10 is 7. We want to find the probability that X is greater
than or equal to 7. The event "at least 7" is the complement of
"less than or equal to 6".

e Using your calculator's distribution menu: 1 — binomcdf(10, .5, 6)
gives 0.171875

e The probability of getting at least 70% of the 10 questions correct
when randomly guessing is approximately 0.172

Exercise:

Problem:

A student takes a 32 question multiple choice exam, but did not study
and randomly guesses each answer. Each question has 3 possible
choices for the answer. Find the probability that the student guesses
more than 75% of the questions correctly.

Solution:

e X =number of questions answered correctly

e X~B(82, 1/3)

e We are interested in MORE THAN 75% of 32 questions correct.
75% of 32 is 24. We want to find P(x>24). The event "more than

24" is the complement of "less than or equal to 24".

e Using your calculator's distribution menu: 1 - binomcdf(32, 1/3,
24)

e P(x>24) = 0.00000026761

e The probability of getting more than 75% of the 32 questions
correct when randomly guessing is very small and practically
zero.

Exercise:

Problem:

Suppose that you are perfoming the probability experiment of rolling
one fair six-sided die. Let F be the event of rolling a"4" or a"5". You
are interested in how many times you need to roll the die in order to
obtain the first “4 or 5” as the outcome.

e p= probability of success (event F occurs)
e q = probability of failure (event F does not occur)

e A Write the description of the random variable X. What are the
values that X can take on? Find the values of p and q.

e B Find the probability that the first occurrence of event F (rolling
a “4” or “5”) is on the second trial.

e C How many trials would you expect until you roll a “4” or “5”?

Solution:

A: X can take on the values 1, 2, 3, .... p = 2/6, gq = 4/6
B20.2222
Crees}

**Exercises 38 - 43 contributed by Roberta Bloom

Review
This module provides a number of homework/review exercises
summarizing topics related to Discrete Random Variables.

The next two questions refer to the following:

A recent poll concerning credit cards found that 35 percent of respondents
use a credit card that gives them a mile of air travel for every dollar they
charge. Thirty percent of the respondents charge more than $2000 per
month. Of those respondents who charge more than $2000, 80 percent use a
credit card that gives them a mile of air travel for every dollar they charge.
Exercise:

Problem:

What is the probability that a randomly selected respondent will spend
more than $2000 AND use a credit card that gives them a mile of air
travel for every dollar they charge?

(0.30) (0.35)
(0.80)(0.35)
(0.80)
(0.80)

0.80)(0.30)
0.80

A
B
C
D

Solution:

C
Exercise:
Problem:
Based upon the above information, are using a credit card that gives a

mile of air travel for each dollar spent AND charging more than $2000
per month independent events?

e AYes
e BNo, and they are not mutually exclusive either
e CNo, but they are mutually exclusive

¢ DNot enough information given to determine the answer

Solution:

B
Exercise:

Problem:

A sociologist wants to know the opinions of employed adult women
about government funding for day care. She obtains a list of 520
members of a local business and professional women’s club and mails
a questionnaire to 100 of these women selected at random. 68
questionnaires are returned. What is the population in this study?

e AAIll employed adult women

e BAI] the members of a local business and professional women’s
club

e CThe 100 women who received the questionnaire

e DAI! employed women with children

Solution:

A

The next two questions refer to the following: An article from The San Jose
Mercury News was concerned with the racial mix of the 1500 students at
Prospect High School in Saratoga, CA. The table summarizes the results.
(Male and female values are approximate.) Suppose one Prospect High
School student is randomly selected.

Ethnic
Group

Gender White Asian Hispanic Black

Male 400 168 115 35
Female 440 132 140 40
Exercise:

American
Indian

16

14

Problem: Find the probability that a student is Asian or Male.

Solution:

0.5773
Exercise:

Problem:

Find the probability that a student is Black given that the student is

Female.

Solution:

0.0522
Exercise:

Problem:

A sample of pounds lost, in a certain month, by individual members of

a weight reducing clinic produced the following statistics:

e Mean = 5 lbs.
e Median = 4.5 lbs.
e Mode = 4 lbs.

e Standard deviation = 3.8 lbs.
e First quartile = 2 lbs.
e Third quartile = 8.5 lbs.

The correct statement is:

e AOne fourth of the members lost exactly 2 pounds.

e BThe middle fifty percent of the members lost from 2 to 8.5 lbs.
¢ CMost people lost 3.5 to 4.5 Ibs.

e¢ DAI! of the choices above are correct.

Solution:

B
Exercise:
Problem:

What does it mean when a data set has a standard deviation equal to
Zero?

e AAIl values of the data appear with the same frequency.
e BThe mean of the data is also zero.

CAIl of the data have the same value.

e DtThere are no data to begin with.

Solution:

C

Exercise:

Problem: The statement that best describes the illustration below is:

e AThe mean is equal to the median.
¢ BThere is no first quartile.

e CThe lowest data value is the median.

e DThe median equals (OF @8)

Solution:

C

Exercise:
Problem:
According to a recent article (San Jose Mercury News) the average
number of babies born with significant hearing loss (deafness) is
approximately 2 per 1000 babies in a healthy baby nursery. The

number climbs to an average of 30 per 1000 babies in an intensive care
nursery.

Suppose that 1000 babies from healthy baby nurseries were randomly
surveyed. Find the probability that exactly 2 babies were born deaf.

Solution:

0.2709
Exercise:
Problem:
A “friend” offers you the following “deal.” For a $10 fee, you may

pick an envelope from a box containing 100 seemingly identical
envelopes. However, each envelope contains a coupon for a free gift.

¢ 10 of the coupons are for a free gift worth $6.
¢ 80 of the coupons are for a free gift worth $8.
¢ 6 of the coupons are for a free gift worth $12.
e 4 of the coupons are for a free gift worth $40.

Based upon the financial gain or loss over the long run, should you
play the game?

e AYes, I expect to come out ahead in money.
e BNo, I expect to come out behind in money.
e ClIt doesn’t matter. I expect to break even.

Solution:

B

The next four questions refer to the following: Recently, a nurse
commented that when a patient calls the medical advice line claiming to
have the flu, the chance that he/she truly has the flu (and not just a nasty
cold) is only about 4%. Of the next 25 patients calling in claiming to have
the flu, we are interested in how many actually have the flu.

Exercise:

Problem: Define the Random Variable and list its possible values.

Solution:

X = the number of patients calling in claiming to have the flu, who
actually have the flu. X = 0, 1, 2, ...25

Exercise:

Problem: State the distribution of X .

Solution:

B(25,0.04)
Exercise:

Problem:

Find the probability that at least 4 of the 25 patients actually have the
flu.

Solution:

0.0165
Exercise:
Problem:

On average, for every 25 patients calling in, how many do you expect
to have the flu?

Solution:

if

The next two questions refer to the following: Different types of writing can
sometimes be distinguished by the number of letters in the words used. A
student interested in this fact wants to study the number of letters of words
used by Tom Clancy in his novels. She opens a Clancy novel at random and
records the number of letters of the first 250 words on the page.

Exercise:

Problem: What kind of data was collected?
e Aqualitative

e Bquantitative - continuous
e Cquantitative — discrete

Solution:

C

Exercise:

Problem: What is the population under study?
Solution:

All words used by Tom Clancy in his novels

Introduction

Continuous Random Variables: Introduction is part of the collection
col10555 written by Barbara Illowsky and Susan Dean and serves as an
introduction to the uniform and exponential distributions with contributions
from Roberta Bloom.

Student Learning Outcomes
By the end of this chapter, the student should be able to:

e Recognize and understand continuous probability density functions in
general.

e Recognize the uniform probability distribution and apply it
appropriately.

e Recognize the exponential probability distribution and apply it
appropriately.

Introduction

Continuous random variables have many applications. Baseball batting
averages, IQ scores, the length of time a long distance telephone call lasts,
the amount of money a person carries, the length of time a computer chip
lasts, and SAT scores are just a few. The field of reliability depends on a
variety of continuous random variables.

This chapter gives an introduction to continuous random variables and the
many continuous distributions. We will be studying these continuous
distributions for several chapters.

Note:The values of discrete and continuous random variables can be
ambiguous. For example, if X is equal to the number of miles (to the
nearest mile) you drive to work, then X is a discrete random variable. You
count the miles. If X is the distance you drive to work, then you measure
values of X and_X is a continuous random variable. How the random
variable is defined is very important.

Properties of Continuous Probability Distributions

The graph of a continuous probability distribution is a curve. Probability is
represented by area under the curve.

The curve is called the probability density function (abbreviated: pdf).
We use the symbol f() to represent the curve. f(x) is the function that
corresponds to the graph; we use the density function f(a) to draw the
graph of the probability distribution.

Area under the curve is given by a different function called the
cumulative distribution function (abbreviated: cdf). The cumulative
distribution function is used to evaluate probability as area.

e The outcomes are measured, not counted.

e The entire area under the curve and above the x-axis is equal to 1.

e Probability is found for intervals of x values rather than for individual
x values.

¢ P(c < x < d) is the probability that the random variable X is in the
interval between the values c and d. P(c < x < d) is the area under
the curve, above the x-axis, to the right of c and the left of d.

e P(x = c) = 0 The probability that x takes on any single individual
value is 0. The area below the curve, above the x-axis, and between
x=c and x=c has no width, and therefore no area (area = 0). Since the
probability is equal to the area, the probability is also 0.

We will find the area that represents probability by using geometry,
formulas, technology, or probability tables. In general, calculus is needed to
find the area under the curve for many probability density functions. When
we use formulas to find the area in this textbook, the formulas were found
by using the techniques of integral calculus. However, because most
students taking this course have not studied calculus, we will not be using
calculus in this textbook.

There are many continuous probability distributions. When using a
continuous probability distribution to model probability, the distribution
used is selected to best model and fit the particular situation.

In this chapter and the next chapter, we will study the uniform distribution,
the exponential distribution, and the normal distribution. The following
graphs illustrate these distributions.

Shaded Area
represents

P(3<x<6)

x

012345678 9 10
The Uniform Distribution

The graph shows a Uniform
Distribution with the area between
x=3 and x=6 shaded to represent
the probability that the value of
the random variable X is in the
interval between 3 and 6.

Shaded Area
represents
P(2<x<4)

123 4 6 6 7 x
The Exponential Distribution

The graph shows an
Exponential Distribution with

the area between x=2 and x=4
shaded to represent the
probability that the value of the
random variable X is in the
interval between 2 and 4.

Shaded area
represents
probability

P(1<x<2)

-3 -2 = 0 1 2 5
The Normal Distribution x

The graph shows the Standard Normal Distribution with
the area between x=1 and x=2 shaded to represent the
probability that the value of the random variable X is in
the interval between 1 and 2.

** With contributions from Roberta Bloom

Glossary

Uniform Distribution
A continuous random variable (RV) that has equally likely outcomes
over the domain, a < x < b. Often referred as the Rectangular
distribution because the graph of the pdf has the form of a rectangle.
Notation: X~U(a,b). The mean is p = ath and the standard deviation

= p\2
iso = / e > Y The probability density function is f(X) = 3 for
a<a«<bora<z < b. The cumulative distribution is

—_ «t~—a
P(X <2#)=—.

Exponential Distribution

A continuous random variable (RV) that appears when we are
interested in the intervals of time between some random events, for
example, the length of time between emergency arrivals at a hospital.
Notation: X~Exp(m). The mean is ps = —- and the standard deviation

is o = +. The probability density function is f(x) = me“™, z > 0
and the cumulative distribution function is P(X < #«) =1—e™.

Continuous Probability Functions

This module introduces the continuous probability function and explores
the relationship between the probability of X and the area under the curve
of f(X).

We begin by defining a continuous probability density function. We use the
function notation f(z). Intermediate algebra may have been your first
formal introduction to functions. In the study of probability, the functions
we study are special. We define the function f(x) so that the area between
it and the x-axis is equal to a probability. Since the maximum probability is
one, the maximum area is also one.

For continuous probability distributions, PROBABILITY = AREA.

Example:
Consider the function f(x) = _ for 0 < x < 20. x = areal number. The
graph of f(x) = = is a horizontal line. However, since 0 < z < 20,

f(a) is restricted to the portion between x = 0 and x = 20, inclusive .

]
f(x) =—
: 20

>

i) — an for 0< x <20.
The graph of f(x) = 36 is a horizontal line segment when 0 < x < 20.

The area between f(x) = ao where 0 < x < 20 and the x-axis is the area
of a rectangle with base = 20 and height =3.

AREA = 20-3 =1

This particular function, where we have restricted x so that the area
between the function and the x-axis is 1, is an example of a continuous
probability density function. It is used as a tool to calculate probabilities.
Suppose we want to find the area between f(x) = — and the x-axis

where 0 < xz < 2.

P|
20 3

NBCU es (0 10))) 2 ee (0)

(2 — 0) = 2 = base of a rectangle

35 = the height.

The area corresponds to a probability. The probability that x is between 0
and 2 is 0.1, which can be written mathematically as

IE se) = EA ee 7))) = (0) IL.

Suppose we want to find the area between f(x) = — and the x-axis
where 4< 24 < 15.

AREA = (15 — 4) - 3 = 0.55

(15 — 4) = 11 = the base of a rectangle

30 = the height.

The area corresponds to the probability P(4 < 2 < 15) = 0.55.
Suppose we want to find P(x =15 ). On an x-y graph, x=15 is a vertical

line. A vertical line has no width (or 0 width). Therefore,
P(x=15) = (base) (height) = (0) (55) = 0.

f(x)

P(X < «) (can be written as P(X < x) for continuous distributions) is
called the cumulative distribution function or CDF. Notice the "less than
or equal to" symbol. We can use the CDF to calculate P(X > x) . The
CDF gives "area to the left" and P(X > x) gives "area to the right." We
calculate P(X > x) for continuous distributions as follows:

VA ie | I ZIO.G ce ta),

f(x)

x
P(X < x) P(X > x) = 1-P(X < x)

Label the graph with f(x) and . Scale the x and y axes with the maximum

x and y values. f(x) = = 0<2 < 20.

f(x)

0 2.3 12.7 x

P(2.3 < @ < 12.7) = (base)(height) = (12.7 — 2.3) (4) = 0.52

Introduction to Normal Distributions

Normal distributions are commonly used in Statistics. While normal
distributions can be quite different, they can all be represented
mathematically and they all have distinct features that will be discussed in
this chapter.

The normal distribution is the most important and widely used distribution
in statistics. It is sometimes called the bell curve although the tonal
qualities of such a bell would be less than pleasing. It is also called the
Gaussian curve after the mathematician Karl-Friedrich Gauss.

Strictly speaking, it is not correct to talk about the normal distribution
since there are many normal distributions. Normal distributions can differ in
their means and in their standard deviations. [link] shows two normal
distributions. The blue distribution has a mean of 50 and a standard
deviation of 10; the distribution in red has a mean of 60 and a standard
deviation of 5. Both distributions are symmetric with relatively more values
at the center of the distribution and relatively few in the tails.

Varieties of Normal Distributions

0 10 20 30 40 50 60 70 80 90 100

Normal distributions
differing in mean and
standard deviation.

The density of the normal distribution (the height for a given value on the x
axis) of the normal distribution is shown below ({link]). The parameters pu
and o are the mean and standard deviation repectively and define the
normal distribution. The symbol e is the base of the natural logarithm and 7
is the constant pi.

Equation:

Since this is a non-mathematical treatment of statistics, do not worry if this
expression confuses you. We will not be referring back to it in later
sections.

Some features of normal distributions are listed below. These features are
illustrated in more detail in the remaining sections of this chapter.

. Normal distributions are symmetric around their mean.

. The mean, median, and mode of a normal distribution are equal.

. The area under the normal curve is equal to 1.0.

. Normal distributions are denser in the center and less dense in the tails.

. Normal distributions are defined by two parameters, the mean (1) and

the standard deviation (0).

6. 68% of the area of a normal distribution is within one standard
deviation of the mean

7.95% of the area of a normal distribution is within two standard
deviations of the mean

8. 99.7% of the area of a normal distribution is within three standard

deviations of the mean

ui BWN RP

The Standard Normal Distribution
This module introduces standard normal distribution and standardizing the
distribution.

As previously discussed, normal distributions do not necessarily have the
Same means and standard deviations. A normal distribution with a mean of
0 and a standard deviation of 1 is called a standard normal distribution
and is typically represented by Z.

Areas underneath the normal distribution are often represented by tables of
the standard normal distribution. A portion of a table of the standard normal
distribution is shown in [link].

Z Area left of Z
-2.50 0.0062
-2.49 0.0064
-2.48 0.0066
-2.47 0.0068
-2.46 0.0069
-2.45 0.0071
-2.44 0.0073
-2.43 0.0075

-2.42 0.0078

Z Area left of Z

-2.41 0.0080
-2.40 0.0082
-2.39 0.0084
-2.38 0.0087
-2.37 0.0089
-2.36 0.0091
-2.35 0.0094
-2.34 0.0096
-2.33 0.0099
-2.32 0.0102

A portion of a table of the standard normal distribution.

The first column titled Z contains values of the standard normal random
variable; the second column contains the area below the curve to the left of
z. Since the distribution has a mean of 0 and a standard deviation of 1, the
Z column is equal to the number of standard deviations below (or above)
the mean. For example, a z of -2.5 represents a value 2.5 standard
deviations below the mean. The area below the curve to the left of z=-2.5
is 0.0062.

The same information can be obtained using a calculator or the following
Java applet. [link] shows how it can be used to compute the area below the
standard normal curve to the left of -2.5. Note that the mean is set to 0 and
the standard deviation is set to 1.

Normal Distribution

Mean: 0 Sd: 1
© Above 1
@seiow [-2.5 |
O Between -1 and 1
O outside -1 or

Shaded area: 0.006210

An example from the applet.

Note:Calculate Areas

A value from any normal distribution can be transformed into its
corresponding value on a standard normal distribution using the following
formula:

where z is the value on the standard normal distribution, x is the value on
the original distribution, pz is the mean of the original distribution and a is
the standard deviation of the original distribution.

As a simple application, what portion of a normal distribution with a mean
of 50 and a standard deviation of 10 is below 26? Applying the formula we
obtain

25 — 50
LS iS CSS

—2.4
10

From [link], we can see that 0.0082 of the distribution is below -2.4. There
is no need to transform to Z if you are using a technology or the applet
shown in [link].

Normal Distribution

10 20 30 40 50 60 70 8d 90

Mean: 50 Sd: 10
© Above 1
@ Below
OC etween -1 and 1
O outside -1 or

Shaded area: 0.008198

Area below 36 in a normal
distribution with a mean of 50 and
a standard deviation of 10.

If all the values in a distribution are transformed to z-scores, then the
distribution will have a mean of 0 and a standard distribution. This process
of transforming a distribution to one with a mean of 0 and a standard
deviation of 1 is called standardizing the distribution.

Calculate Areas

Applet failed to run. No Java plug-in was found.

Glossary

standard normal distribution
The standard normal distribution is a normal distribution with a
mean of 0 and a standard deviation of 1.

Z-Scores

If X is a normally distributed random variable and X~N(u, o), then the z-
score is:
Equation:

The z-score tells you how many standard deviations that the value z is
above (to the right of) or below (to the left of) the mean, jz. Values of z
that are larger than the mean have positive z-scores and values of x that are
smaller than the mean have negative z-scores. If x equals the mean, then x
has a z-score of 0.

Example:

Suppose X ~ N(5, 6). This says that X is a normally distributed random
variable with mean pp = 5 and standard deviation o = 6. Suppose x = 17.
Then:

Equation:

This means that x = 17 is 2 standard deviations (20) above or to the
right of the mean pt = 5. The standard deviation is o = 6.

Notice that:

Equation:

Do 2 16 —vls, (The pattern is + zo = 2.)

Now suppose x=1. Then:
Equation:

= ie
= —- = —0.67 (rounded to two decimal places)
o

This means that x = 1 is 0.67 standard deviations (- 0.670) below or to
the left of the mean pi = 5. Notice that:

5 + (—0.67)(6) is approximately equal to 1 (This has the pattern
p+ (—0.67)o = 1)

Summarizing, when z is positive, x is above or to the right of and when
zis negative, x is to the left of or below p.

Example:

Some doctors believe that a person can lose 5 pounds, on the average, in a
month by reducing his/her fat intake and by exercising consistently.
Suppose weight loss has a normal distribution. Let X = the amount of
weight lost (in pounds) by a person in a month. Use a standard deviation of
2 pounds. X~N(5, 2). Fill in the blanks.

Exercise:

Problem:

Suppose a person lost 10 pounds in a month. The z-score when
x = 10 pounds is z = 2.5 (verify). This z-score tells you that x = 10

is standard deviations to the (right or left) of the
mean (What is the mean?).
Solution:

This z-score tells you that x = 10 is 2.5 standard deviations to the
right of the mean 5.

Exercise:
Problem:
Suppose a person gained 3 pounds (a negative weight loss). Then z =

. This z-score tells you that x = -3 is standard
deviations to the (right or left) of the mean.

Solution:

= -4. This z-score tells you that x = -3 is 4 standard deviations to
the left of the mean.
Suppose the random variables X and Y have the following normal
distributions: X ~N(5, 6) and Y ~ N(2, 1). Ifx = 17, then z = 2. (This
was previously shown.) If y = 4, what is z?
Equation:

= SS where p}=2 and o=1.

The z-score for y = 4 is z = 2. This means that 4 is z = 2 standard
deviations to the right of the mean. Therefore, x = 17 and y = 4 are both 2
(of their) standard deviations to the right of their respective means.

The z-score allows us to compare data that are scaled differently. To
understand the concept, suppose X ~N(5, 6) represents weight gains for
one group of people who are trying to gain weight in a 6 week period and
Y ~N(2, 1) measures the same weight gain for a second group of people.
A negative weight gain would be a weight loss. Since x = 17 and y = 4
are each 2 standard deviations to the right of their means, they represent
the same weight gain relative to their means.

The Empirical Rule
If X is arandom variable and has a normal distribution with mean p and
standard deviation o then the Empirical Rule says (See the figure below)

e About 68.27% of the x values lie between -1o0 and +10 of the mean p
(within 1 standard deviation of the mean).

e About 95.45% of the x values lie between -20 and +20 of the mean p
(within 2 standard deviations of the mean).

e About 99.73% of the x values lie between -30 and +30 of the mean p
(within 3 standard deviations of the mean). Notice that almost all the x
values lie within 3 standard deviations of the mean.

e The z-scores for +1o and —1o are +1 and -1, respectively.

e The z-scores for +20 and —2o0 are +2 and -2, respectively.

e The z-scores for +30 and —3a0 are +3 and -3 respectively.

xX
—30-—20-lo wp lo 20 30

The Empirical Rule is also known as the 68-95-99.7 Rule.

Example:

Suppose X has a normal distribution with mean 50 and standard deviation
6.

e About 68.27% of the x values lie between -1o = (-1)(6) = -6 and lo =
(1)(6) = 6 of the mean 50. The values 50 - 6 = 44 and 50 + 6 = 56 are
within 1 standard deviation of the mean 50. The z-scores are -1 and
+1 for 44 and 56, respectively.

e About 95.45% of the x values lie between -20 = (-2)(6) = -12 and 20
= (2)(6) = 12 of the mean 50. The values 50 - 12 = 38 and 50 + 12 =
62 are within 2 standard deviations of the mean 50. The z-scores are
-2 and 2 for 38 and 62, respectively.

e About 99.73% of the x values lie between -3o0 = (-3)(6) = -18 and 30
= (3)(6) = 18 of the mean 50. The values 50 - 18 = 32 and 50 + 18 =
68 are within 3 standard deviations of the mean 50. The z-scores are
-3 and +3 for 32 and 68, respectively.

Normal Distribution: Areas to the Left and Right of x
The arrow in the graph below points to the area to the left of x. This area is

represented by the probability P(X < x). Normal tables, computers, and
calculators provide or calculate the probability P(X < 2).

P(X <x)

X
X

The area to the right is then P(X > x) = 1— P(X <2).
Remember, P(X < x) = Area to the left of the vertical line through z.

P(X >a) =1-— P(X < a) =. Area to the right of the vertical line
through x

P(X < 2) is the same as P(X < x) and P(X > z) is the same as
P(X > a) for continuous distributions.

Areas of Normal Distributions

Areas under portions of a normal distribution can be computed by using
calculus. Since this is a non-mathematical treatment of statistics, we will
rely on computer programs and tables to determine these areas. [link] shows
a normal distribution with a mean of 50 and a standard deviation of 10. The
shaded area between 40 and 60 contains 68% of the distribution.

Normal! distribution with a
mean of 50 and standard
deviation of 10. 68% of the
area is within one standard
deviation (10) of the mean
(50).

[link] shows a normal distribution with a mean of 100 and a standard
deviation of 20. As in Figure 1, 68% of the distribution is within one
standard deviation of the mean.

Normal distribution
with a mean of 100
and standard
deviation of 20. 68%
of the area is within

one standard
deviation (20) of the
mean (100).

The normal distributions shown in [link] and [link] are specific examples of
the general rule that 68% of the area of any normal distribution is within
one standard deviation of the mean.

Note: 68% of the area of any normal distribution is within one standard
deviation of the mean

[link] shows a normal distribution with a mean of 75 and a standard
deviation of 10. The shaded area contains 95% of the area and extends from
55.4 to 94.6. For all normal distributions, 95% of the area is within 1.96
standard deviations of the mean. For quick approximations, it is sometimes
useful to round off and use 2 rather than 1.96 as the number of standard
deviations you need to extend from the mean so as to include 95% of the
area.

A normal distribution
with a mean of 75 and
a standard deviation
of 10. 95% of the area
is within 1.96
standard deviations of
the mean.

The Java applet "Calculate Area for a given X " can be used to calculate
areas under the normal distribution. Use it to find the proportion of a
normal distribution with a mean of 90 and a standard deviation of 12 that is
above 110. Set the mean to 90 and the standard deviation to 12. Then enter
"110" in the box to the right of the radio button "Above." At the bottom of
the display you will see that the shaded area is 0.04779. See if you can use
the applet to find that the area between 115 and 120 is 0.012401.

Bem Tort eon

{ a.

‘,
‘

/
/
/
JZ
Pail

2 = te 3 92 Ws 1.4 Lee 138

Mean: 9) $e: L2
® sbeve
aziow 1

1
1

£3 aotwcer ard 1

©) aurside oo

Shaded area: 2.247792

Display from
applet showing
the area above

110.

The applet "Calculate X for a given Area" works in reverse. For example,
say you wanted to find the score corresponding to the 75th percentile of a
normal distribution with a mean of 90 and a standard deviation of 12. You
enter 90 for the mean and 12 for the standard deviation. Then, enter 0.75 for

the shaded area and click the "Below" button. The area below 98.0939 is
O75.

Horna Usticeties

Mean: 69 | 9 Sd: 9D
Shaded Arse:
Above
® bei: 99.2929
Bemween
Ouside
Display from applet
showing that the

7oth percentile is
98.093.

Calculations of Probabilities

Probabilities are calculated by using technology. There are instructions in
the chapter for the TI-83+ and TI-84 calculators.

Note:In the Table of Contents for Collaborative Statistics, entry 15.
Tables has a link to a table of normal probabilities. Use the probability
tables if so desired, instead of a calculator. The tables include instructions
for how to use then.

Example:
If the area to the left is 0.0228, then the area to the right is
1 — 0.0228 = 0.9772.

Example:

The final exam scores in a statistics class were normally distributed with a
mean of 63 and a standard deviation of 5.

Exercise:

Problem:

Find the probability that a randomly selected student scored more than
65 on the exam.

Solution:

Let X = ascore on the final exam. X~N(63, 5), where ps = 63 and
(5)

Draw a graph.

Then, find P(« > 65).

P(x > 65) = 0.3446 (calculator or computer)

0.3446

a

The probability that one student scores more than 65 is 0.3446.

63 65

Using the TI-83+ or the TI-84 calculators, the calculation is as
follows. Go into 2nd DISTR.

After pressing 2nd DISTR, press 2:normalcdf.
The syntax for the instructions are shown below.

normalcdf(lower value, upper value, mean, standard deviation) For
this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 ( =
10°°) by pressing 1, the EE key (a 2nd key) and then 99. Or, you can
enter 10/99 instead. The number 10% is way out in the right tail of
the normal curve. We are calculating the area between 65 and 10°”. In
some instances, the lower number of the area might be -1E99 ( =
—10%). The number —10% is way out in the left tail of the normal
curve.

Note:The TI probability program calculates a z-score and then the
probability from the z-score. Before technology, the z-score was
looked up in a standard normal probability table (because the math
involved is too cumbersome) to find the probability. In this example,

a standard normal table with area to the left of the z-score was used.
You calculate the z-score and look up the area to the left. The
probability is the area to the right.

ea S88 — 0.4 . Area to the left is 0.6554.
(pe Oa) 0 le ood Ur ado

Exercise:

Problem:

Find the probability that a randomly selected student scored less than
85.

Solution:

Draw a graph.

Then find P(x < 85). Shade the graph. P(a < 85) = 1 (calculator
or computer)

The probability that one student scores less than 85 is approximately 1
(or 100%).

The TI-instructions and answer are as follows:

normalcdf(0,85,63,5) = 1 (rounds to 1)
Exercise:

Problem:

Find the 90th percentile (that is, find the score k that has 90 % of the
scores below k and 10% of the scores above k).

Solution:

Find the 90th percentile. For each problem or part of a problem, draw
a new graph. Draw the x-axis. Shade the area that corresponds to the
90th percentile.

Let k = the 90th percentile. & is located on the x-axis. P(x < k) is
the area to the left of k. The 90th percentile k separates the exam
scores into those that are the same or lower than & and those that are
the same or higher. Ninety percent of the test scores are the same or
lower than & and 10% are the same or higher. k is often called a
critical value.

k = 69.4 (calculator or computer)

P(x < k) = 0.90

The 90th percentile is 69.4. This means that 90% of the test scores fall
at or below 69.4 and 10% fall at or above. For the TI-83+ or TI-84
calculators, use LnvNormin 2nd DISTR. invNorm(area to the left,
mean, standard deviation) For this problem, invNorm(0.90,63,5) =
69.4

Exercise:

Problem:

Find the 70th percentile (that is, find the score k such that 70% of
scores are below k and 30% of the scores are above k).

Solution:

Find the 70th percentile.
Draw a new graph and label it appropriately. k = 65.6

The 70th percentile is 65.6. This means that 70% of the test scores fall
at or below 65.5 and 30% fall at or above.

invNorm(0.70,63,5) = 65.6

Example:

A computer is used for office work at home, research, communication,
personal finances, education, entertainment, social networking and a
myriad of other things. Suppose that the average number of hours a
household personal computer is used for entertainment is 2 hours per day.
Assume the times for entertainment are normally distributed and the
standard deviation for the times is half an hour.

Exercise:

Problem:

Find the probability that a household personal computer is used
between 1.8 and 2.75 hours per day.

Solution:

Let X = the amount of time (in hours) a household personal computer
is used for entertainment. x~N(2, 0.5) where « = 2 and o = 0.5.

Find P(1.8 < x < 2.75).

The probability for which you are looking is the area between
Les AMC aN) peel Out 2 id) OOO

18 5 2.75. Xx

normalcdf(1.8,2.75,2,0.5) = 0.5886
The probability that a household personal computer is used between
1.8 and 2.75 hours per day for entertainment is 0.5886.
Exercise:
Problem:

Find the maximum number of hours per day that the bottom quartile
of households use a personal computer for entertainment.

Solution:

To find the maximum number of hours per day that the bottom
quartile of households uses a personal computer for entertainment,
find the 25th percentile, k, where P(x < k) = 0.25.

k= 1.67

P(x > k) = 0.75
P(x <k) = 0.25 ia

invNorm(0.25,2,.5) = 1.66

The maximum number of hours per day that the bottom quartile of
households uses a personal computer for entertainment is 1.66 hours.

Summary of Formulas
Formula
Normal Probability Distribution

X~N(p,0)

j = the mean o = the standard deviation
Formula
Standard Normal Probability Distribution

Z~N(0, 1)
z = a Standardized value (z-score)

mean = 0 standard deviation = 1
Formula
Finding the kth Percentile

To find the kth percentile when the z-score is known: k = w+ (z)o
Formula
z-score
o—U
on
Formula

Finding the area to the left

—

The area to the left: P(X < x)
Formula
Finding the area to the right

The area to the right: P(X > 7) = 1— P(X < 2)

Practice: The Normal Distribution

Student Learning Outcomes

e The student will analyze data following a normal distribution.

Given

The life of Sunshine CD players is normally distributed with a mean of 4.1
years and a standard deviation of 1.3 years. A CD player is guaranteed for 3
years. We are interested in the length of time a CD player lasts.

Normal Distribution
Exercise:

Problem: Define the Random Variable X in words. X =

Exercise:

Problem: X~
Exercise:

Problem:

Find the probability that a CD player will break down during the
guarantee period.

e a Sketch the situation. Label and scale the axes. Shade the region
corresponding to the probability.

ebP(0<a< = (Use zero (0) for the
minimum value of x.)

Solution:
¢ b3,0.1979
Exercise:
Problem:
Find the probability that a CD player will last between 2.8 and 6 years.

e a Sketch the situation. Label and scale the axes. Shade the region
corresponding to the probability.

‘bP(._. <a< =

Solution:
e b 2.8,6,0.7694
Exercise:
Problem:

Find the 70th percentile of the distribution for the time a CD player
lasts.

e a Sketch the situation. Label and scale the axes. Shade the region
corresponding to the lower 70%.

___——, Therefore, k =

© bP(4#<k)=

Solution:

¢ b0.70,4.78years

Homework
Exercise:

Problem:

According to a study done by De Anza students, the height for Asian
adult males is normally distributed with an average of 66 inches and a
standard deviation of 2.5 inches. Suppose one Asian adult male is
randomly chosen. Let X =height of the individual.

e aX~ ( ; )

e b Find the probability that the person is between 65 and 69
inches. Include a sketch of the graph and write a probability
statement.

e¢ c Would you expect to meet many Asian adult males over 72
inches? Explain why or why not, and justify your answer
numerically.

e d The middle 40% of heights fall between what two values?
Sketch the graph and write the probability statement.

Solution:

e a N(66,2.5)

e b0.5404

e cNo

e dBetween 64.7 and 67.3 inches

Exercise:

Problem:

IQ is normally distributed with a mean of 100 and a standard deviation
of 15. Suppose one individual is randomly chosen. Let X =IQ of an
individual.

e a X~ ( : )
e b Find the probability that the person has an IQ greater than 120.
Include a sketch of the graph and write a probability statement.

¢ c Mensa is an organization whose members have the top 2% of all
IQs. Find the minimum IQ needed to qualify for the Mensa
organization. Sketch the graph and write the probability
statement.

e d The middle 50% of IQs fall between what two values? Sketch
the graph and write the probability statement.

Exercise:

Problem:

The percent of fat calories that a person in America consumes each
day is normally distributed with a mean of about 36 and a standard
deviation of 10. Suppose that one individual is randomly chosen. Let
X =percent of fat calories.

e a X~ ( F )

e b Find the probability that the percent of fat calories a person
consumes is more than 40. Graph the situation. Shade in the area
to be determined.

e c Find the maximum number for the lower quarter of percent of
fat calories. Sketch the graph and write the probability statement.

Solution:

¢ a N(36,10)
° b 0.3446
© ¢ 29.3

Exercise:
Problem:
Suppose that the distance of fly balls hit to the outfield (in baseball) is

normally distributed with a mean of 250 feet and a standard deviation
of 50 feet.

e alf X = distance in feet for a fly ball, then X~
( )

¢ b If one fly ball is randomly chosen from this distribution, what is
the probability that this ball traveled fewer than 220 feet? Sketch
the graph. Scale the horizontal axis X. Shade the region
corresponding to the probability. Find the probability.

e c Find the 80th percentile of the distribution of fly balls. Sketch
the graph and write the probability statement.

Exercise:

Problem:

In China, 4-year-olds average 3 hours a day unsupervised. Most of the
unsupervised children live in rural areas, considered safe. Suppose that
the standard deviation is 1.5 hours and the amount of time spent alone
is normally distributed. We randomly survey one Chinese 4-year-old
living in a rural area. We are interested in the amount of time the child
spends alone per day. (Source: San Jose Mercury News)

e alin words, define the random variable X.X =

© bxX~

e c Find the probability that the child spends less than 1 hour per
day unsupervised. Sketch the graph and write the probability
statement.

e d What percent of the children spend over 10 hours per day
unsupervised?

e e 70% of the children spend at least how long per day
unsupervised?

Solution:

e a the time (in hours) a 4-year-old in China spends unsupervised
per day

¢ b N(3,1.5)

e c 0.0912

e d0

e 2.21 hours

Exercise:

Problem:

In the 1992 presidential election, Alaska’s 40 election districts
averaged 1956.8 votes per district for President Clinton. The standard
deviation was 572.3. (There are only 40 election districts in Alaska.)
The distribution of the votes per district for President Clinton was bell-
shaped. Let X = number of votes for President Clinton for an election
district. (Source: The World Almanac and Book of Facts)

a State the approximate distribution of X. X~

b Is 1956.8 a population mean or a sample mean? How do you
know?

c Find the probability that a randomly selected district had fewer
than 1600 votes for President Clinton. Sketch the graph and write
the probability statement.

d Find the probability that a randomly selected district had
between 1800 and 2000 votes for President Clinton.

e Find the third quartile for votes for President Clinton.

Exercise:

Problem:

Suppose that the duration of a particular type of criminal trial is known
to be normally distributed with a mean of 21 days and a standard
deviation of 7 days.

a In words, define the random variable X. X =

b X~

c If one of the trials is randomly chosen, find the probability that
it lasted at least 24 days. Sketch the graph and write the
probability statement.

d 60% of all of these types of trials are completed within how
many days?

Solution:

e aThe duration of a criminal trial
¢ b N(21,7)

¢ ¢ 0.3341

e d 22.77

Exercise:

Problem:

Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per
2.5 mile lap (in a 7 lap race) with a standard deviation of 2.28 seconds
. The distribution of her race times is normally distributed. We are
interested in one of her randomly selected laps. (Source: log book of
Terri Vogel)

e a ln words, define the random variable X.X =
© bxXe
e c Find the percent of her laps that are completed in less than 130
seconds.
e d The fastest 3% of her laps are under ;
e e The middle 80% of her laps are from seconds to
seconds.

Exercise:

Problem:

Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as
to how long customers at Lucky claimed to wait in the checkout line
until their turn. Let _X =time in line. Below are the ordered real data
(in minutes):

0.50 4.25 2) 6 720

1.75 4.25 5.25 6 peas
4.25 D205 6.25 Ao
2:25 4.25 ops) 6.25 719
229 4.5 5.0 6.5 8
2.5 4.75 5.0 6.5 8.25
2:15 4.75 D./0 6.5 9.5
3.20 4.75 9.75 6.75 9.5
3.75 fs) 6 6.75 O37)
3.75 fs) 6 6.75 10.75

a Calculate the sample mean and the sample standard deviation.
b Construct a histogram. Start the x — axis at —0.375 and make
bar widths of 2 minutes.

c Draw a smooth curve through the midpoints of the tops of the
bars.

d In words, describe the shape of your histogram and smooth
curve.

e Let the sample mean approximate pz and the sample standard
deviation approximate o. The distribution of X can then be
approximated by X~

f Use the distribution in (e) to calculate the probability that a
person will wait fewer than 6.1 minutes.

g Determine the cumulative relative frequency for waiting less
than 6.1 minutes.

h Why aren’t the answers to (f) and (g) exactly the same?

i Why are the answers to (f) and (g) as close as they are?

e j If only 10 customers were surveyed instead of 50, do you think
the answers to (f) and (g) would have been closer together or
farther apart? Explain your conclusion.

Solution:

e a The sample mean is 5.51 and the sample standard deviation is
215

e e N(5.51,2.15)

¢ f 0.6081

e g 0.64

Exercise:

Problem:

Suppose that Ricardo and Anita attend different colleges. Ricardo’s
GPA is the same as the average GPA at his school. Anita’s GPA is 0.70
standard deviations above her school average. In complete sentences,
explain why each of the following statements may be false.

e a Ricardo’s actual GPA is lower than Anita’s actual GPA.
e b Ricardo is not passing since his z-score is zero.
e c Anita is in the 70th percentile of students at her college.

Exercise:

Problem:

Below is a sample of the maximum capacity (maximum number of
spectators) of sports stadiums. The table does not include horse racing
or motor racing stadiums. (Source:
http://en.wikipedia.org/wiki/List_of_stadiums_by_capacity)

40,000
49,133
51,500
52,692
59,000
59,680
62,872
66,161
70,585

75,025

40,000
50,071
51,900
53,864
59,000
60,000
64,035
67,428
71,594

76,212

45,050
50,096
52,000
54,000
55,000
60,000
65,000
68,349
72,000

78,000

45,500
50,466
52,132
55,000
55,082
60,492
65,050
68,976
72,922

80,000

46,249
50,832
52,200
59,000
57,000
60,580
65,647
69,372
735379

80,000

48,134
51,100
52,530
55,000
58,008
62,380
66,000
70,107
74,500

82,300

e a Calculate the sample mean and the sample standard deviation

for the maximum capacity of sports stadiums (the data).

e b Construct a histogram of the data.
e c Draw a smooth curve through the midpoints of the tops of the

bars of the histogram.

e dIn words, describe the shape of your histogram and smooth

curve.

e e Let the sample mean approximate p and the sample standard
deviation approximate o. The distribution of X can then be

approximated by X~

e f Use the distribution in (e) to calculate the probability that the

maximum capacity of sports stadiums is less than 67,000

spectators.
e g Determine the cumulative relative frequency that the maximum
capacity of sports stadiums is less than 67,000 spectators. Hint:

Order the data and count the sports stadiums that have a

maximum capacity less than 67,000. Divide by the total number
of sports stadiums in the sample.
e h Why aren’t the answers to (f) and (g) exactly the same?

Solution:

e a The sample mean is 60,136.4 and the sample standard deviation
is 10,468.1.

e e N(60136.4,10468.1)

¢ f 0.7440

e« g 0.7167

Try These Multiple Choice Questions

The questions below refer to the following: The patient recovery time
from a particular surgical procedure is normally distributed with a mean of
5.3 days and a standard deviation of 2.1 days.

Exercise:

Problem: What is the median recovery time?

© A2.7
e B53
e C74
e D2.1

Solution:

B
Exercise:

Problem:

What is the z-score for a patient who takes 10 days to recover?

Al.5
B 0.2
C 2.2
D 7.3

Solution:

C
Exercise:

Problem:
What is the probability of spending more than 2 days in recovery?

e A 0.0580
e B 0.8447
e € 0.0553
e D 0.9420

Solution:

D

Exercise:

Problem: The 90th percentile for recovery times is?

e A8.89
¢ B7.07
e C 7.99
e D432

Solution:

@

The questions below refer to the following: The length of time to find a
parking space at 9 A.M. follows a normal distribution with a mean of 5
minutes and a standard deviation of 2 minutes.

Exercise:

Problem:

Based upon the above information and numerically justified, would
you be surprised if it took less than 1 minute to find a parking space?

e A Yes
e BNo
e C Unable to determine

Solution:

A
Exercise:

Problem:

Find the probability that it takes at least 8 minutes to find a parking
space.

e A 0.0001
¢ B 0.9270
e C 0.1862
e D 0.0668

Solution:

D

Exercise:

Problem:

Seventy percent of the time, it takes more than how many minutes to
find a parking space?

© A1.24
e B2.41
e €3.95
e D6.05

Solution:

G
Exercise:
Problem:

If the mean is significantly greater than the standard deviation, which
of the following statements is true?

e I The data cannot follow the uniform distribution.
e II The data cannot follow the exponential distribution..
e III The data cannot follow the normal distribution.

¢ Alonly

¢ BI only

¢ C Ill only

e DI, Ul, and Ill

Solution:

B

Review

The next two questions refer to: X ~ U(3, 13)
Exercise:

Problem: Explain which of the following are false and which are true.

° a f(z) = 75,3 <2< 13
e b There is no mode.

e c The median is less than the mean.
e dP(x > 10) = P(x < 6)

Solution:

e a True

e b True

e c False — the median and the mean are the same for this
symmetric distribution

e d True

Exercise:

Problem: Calculate:

e a Mean
e b Median
e c 65th percentile.

Solution:

e a8

°b8
°c P(x < k) =0.65 = (k—3) *(4).k=9.5

Exercise:

Problem: Which of the following is true for the above box plot?

e a 25% of the data are at most 5.

¢ b There is about the same amount of data from 4 — 5 as there is
from 5— 7.

e c There are no data values of 3.

e d 50% of the data are 4.

Solution:

e a False — + of the data are at most 5

e b True — each quartile has 25% of the data
e c False — that is unknown
e d False — 50% of the data are 4 or less

Exercise:

Problem:
If P(G | H) = P(G)), then which of the following is correct?

e AG and H are mutually exclusive events.

¢ BP(G) = P(A)

e C Knowing that H has occurred will affect the chance that G will
happen.

¢ DG and H are independent events.

Solution:

D

Exercise:

Problem:

If P(J) = 0.3, P(K) = 0.6, and J and K are independent events,
then explain which are correct and which are incorrect.

© AP
° BP
© CP
« DP

Jand Kk) =0
Jor kK) = 0,9
J or Kk) = 0.72
J) # PJ | K)

LN LN NON

Solution:

e A False - J and K are independent so they are not mutually
exclusive which would imply dependency (meaning P(J and K) is
not 0).

e B False - see answer C.

e C True - P(J or K) = P(J) + P(K) - P(J and K) = P(J) + P(K) -
P(J)P(K) = 0.3 + 0.6 - (0.3)(0.6) = 0.72. Note that P(J and K) =
P(J)P(K) because J and K are independent.

¢ D False - J and K are independent so P(J) = P(J|K).

Exercise:

Problem:

On average, 5 students from each high school class get full
scholarships to 4-year colleges. Assume that most high school classes
have about 500 students.

X =the number of students from a high school class that get full
scholarships to 4-year school. Which of the following is the
distribution of X?

e AP(5)
¢ BB(500,5)
e CExp(1/5)

¢ DN(5, (0.01)(0.99)/500)

Solution:

A

Introduction
This module provides a brief introduction to the Central Limit Theorem.

Student Learning Outcomes
By the end of this chapter, the student should be able to:

e Recognize the Central Limit Theorem problems.

¢ Classify continuous word problems by their distributions.
e Apply and interpret the Central Limit Theorem for Means.
e Apply and interpret the Central Limit Theorem for Sums.

Introduction

Why are we so concerned with means? Two reasons are that they give us a
middle ground for comparison and they are easy to calculate. In this
chapter, you will study means and the Central Limit Theorem.

The Central Limit Theorem (CLT for short) is one of the most powerful
and useful ideas in all of statistics. Both alternatives are concerned with
drawing finite samples of size n from a population with a known mean, pL,
and a known standard deviation, o. The first alternative says that if we
collect samples of size n and n is "large enough," calculate each sample's
mean, and create a histogram of those means, then the resulting histogram
will tend to have an approximate normal bell shape. The second alternative
says that if we again collect samples of size n that are "large enough,"
calculate the sum of each sample and create a histogram, then the resulting
histogram will again tend to have a normal bell-shape.

In either case, it does not matter what the distribution of the original
population is, or whether you even need to know it. The important fact
is that the sample means and the sums tend to follow the normal
distribution. And, the rest you will learn in this chapter.

The size of the sample, n, that is required in order to be to be ‘large enough’
depends on the original population from which the samples are drawn. If
the original population is far from normal then more observations are

needed for the sample means or the sample sums to be normal. Sampling is
done with replacement.

Optional Collaborative Classroom Activity

Do the following example in class: Suppose 8 of you roll 1 fair die 10
times, 7 of you roll 2 fair dice 10 times, 9 of you roll 5 fair dice 10 times,
and 11 of you roll 10 fair dice 10 times.

Each time a person rolls more than one die, he/she calculates the sample
mean of the faces showing. For example, one person might roll 5 fair dice
and get a 2, 2, 3, 4, 6 on one roll.

The mean is The 3.4 is one mean when 5 fair dice

242434446 _ 9 4
4,
are rolled. This same person would roll the 5 dice 9 more times and

calculate 9 more means for a total of 10 means.

Your instructor will pass out the dice to several people as described above.
Roll your dice 10 times. For each roll, record the faces and find the mean.
Round to the nearest 0.5.

Your instructor (and possibly you) will produce one graph (it might be a
histogram) for 1 die, one graph for 2 dice, one graph for 5 dice, and one
graph for 10 dice. Since the "mean" when you roll one die, is just the face
on the die, what distribution do these means appear to be representing?

Draw the graph for the means using 2 dice. Do the sample means show
any kind of pattern?

Draw the graph for the means using 5 dice. Do you see any pattern
emerging?

Finally, draw the graph for the means using 10 dice. Do you see any
pattern to the graph? What can you conclude as you increase the number of
dice?

As the number of dice rolled increases from 1 to 2 to 5 to 10, the following
is happening:

1. The mean of the sample means remains approximately the same.

2. The spread of the sample means (the standard deviation of the sample
means) gets smaller.

3. The graph appears steeper and thinner.

You have just demonstrated the Central Limit Theorem (CLT).

The Central Limit Theorem tells you that as you increase the number of
dice, the sample means tend toward a normal distribution (the
sampling distribution).

Glossary

Average
A number that describes the central tendency of the data. There are a
number of specialized averages, including the arithmetic mean,
weighted mean, median, mode, and geometric mean.

Central Limit Theorem
Given a random variable (RV) with known mean pu and known
standard deviation o. We are sampling with size n and we are
interested in two new RVs - the sample mean, X, and the sample sum,
3/X. If the size n of the sample is sufficiently large, then X ~

N(n, <) and XX ~ N(np, no). If the size n of the sample is

sufficiently large, then the distribution of the sample means and the
distribution of the sample sums will approximate a normal distribution
regardless of the shape of the population. The mean of the sample
means will equal the population mean and the mean of the sample
sums will equal n times the population ea The standard deviation
of the distribution of the sample means, Tn , is called the standard

error of the mean.

Introduction to Sampling Distributions

Suppose you randomly sampled 10 people from the population of women in
Houston Texas between the ages of 21 and 35 years and computed the mean
height of your sample. You would not expect your sample mean to be equal
to the mean of all women in Houston. It might be somewhat lower or it
might be somewhat higher, but it would not equal the population mean
exactly. Similarly, if you took a second sample of 10 people from the same
population, you would not expect the mean of this second sample to equal
the mean of the first sample.

Recall that inferential statistics concerns generalizing from a sample to a
population. A critical part of inferential statistics involves determining how
far sample statistics are likely to vary from each other and from the
population parameter. (In this example, the sample means are sample
Statistics and the sample parameter is the population mean.) As the later
portions of this chapter show (Sampling Distribution of the Mean and
Sampling Distribution of Difference Between Means), these determinations
are based on sampling distributions.

Discrete Distributions

We will illustrate the concept of sampling distributions with a simple
example. [link] shows three pool balls, each with a number on it. Two of the
balls are selected randomly (with replacement) and the average of their
numbers is computed.

The pool balls.

All possible outcomes are shown in [link].

Outcome Ball 1 Ball 2 Mean
1 1 1 1.0
2 1 2 1.5
3 1 3 2.0
4 Z 1 1.5
5 2 2 2.0
6 2 3 25
7 3 1 2.0
8 3 2 2.5
9 3 3 3.0

All possible outcomes when two balls are sampled

Notice that all the means are either 1.0, 1.5, 2.0, 2.5, or 3.0. The frequencies
of these means are shown in [link]. The relative frequencies are equal to the
frequencies divided by nine because there are nine possible outcomes.

Mean Frequency Relative Frequency

1.0 1 0.111
15 2 0.222
2.0 3 0.333
25 2 0.222
3.0 1 0.111

Frequencies of means for N = 2.

[link] shows a relative frequency distribution of the means based on
[link]. This distribution is also a probability distribution since the Y-axis
is the probability of obtaining a given mean from a sample of two balls in
addition to being the relative frequency.

0,350

0,300

0,250

0,200

0,150

0.100

Realtive Frequency (Probability)

0,050

0,000
I 15 2 25 3

Distribution of means for N = 2.

The distribution shown in [link] is called the sampling distribution of the
mean. Specifically, it is the sampling distribution of the mean for a sample
size of 2( N = 2 ). For this simple example, the distribution of pool balls
and the sampling distribution are both discrete distribution. The pool balls
have only the numbers 1, 2, and 3, and a sample mean can have one of only
five possible values.

There is an alternative way of conceptualizing a sampling distribution that
will be useful for more complex distributions. Imagine that two balls are
sampled (with replacement) and the mean of the two balls is computed and
recorded. Then this process is repeated for a second sample, a third sample,
and eventually thousands of a samples. After thousands of samples are
taken and the mean computed for each, a relative frequency distribution is
drawn. The more samples, the closer the relative frequency distribution will

come to the sampling distribution shown in [link]. As the number of
samples approaches infinity, the frequency distribution will approach the
sampling distribution. This means that you can conceive of a sampling
distribution as being a frequency distribution based on a very large number
of samples. To be strictly correct, the sampling distribution only equals the
frequency distribution exactly when there is an infinite number of samples.

It is important to keep in mind that every statistic, not just the mean has a
sampling distribution. For example, [link] shows all possible outcomes for
the range of two numbers (larger number minus the smaller number). [link]
shows the frequencies for each of the possible ranges and [link] shows the
sampling distribution of the range.

Outcome Ball 1 Ball 2 Range
1 1 1 0
2 1 2 1
3 1 3 2
4 2 1 1
5 2 2 0
6 2 3 1
7 3 1 2
8 3 2 1

All possible outcomes when two balls are sampled.

Range

2

Frequency
3
4

2

Frequencies of ranges for N = 2.

0,500

0.450

o 9 9 9 9
No th QQ im) -
o on o on o
o o o o o

Realtive Frequency (Probability)

° °
am
3 oO
3 o

0,050

0,000

Relative Frequency
0.333
0.444

0.222

Distribution of ranges for N = 2.

It is also important to keep in mind that there is a sampling distribution for
various sample sizes. For simplicity, we have been using VV = 2. The
sampling distribution of the range for N = 3 is shown in [link].

0,500

0.450

o
=
o
o

Oo oO
i o3
p=] on
o o

ar)
Nh

8 &
o o

Realtive Frequency (Probability)

o 9
as as
o on
o o

0,050

0,000

0 1 2
Range

Distribution of ranges for N = 3.

Continuous Distributions

In the previous section, the population consisted of three pool balls. Now
we will consider sampling distributions when the population distribution is
continuous. What if we had a thousand pool balls with numbers ranging
from 0.001 to 1.000 in equal steps. (Although this distribution is not really
continuous, it is close enough to be considered continuous for practical
purposes.) As before, we are interested in the distribution of means we
would get if we sampled two balls and computed the mean of these two. In
the previous example, we started by computing the mean for each of the
nine possible outcomes. This would get a bit tedious for this problem since
there are 1,000,000 possible outcomes (1,000 for the first ball x 1,000 for
the second.) Therefore, it is more convenient to use our second
conceptualization of sampling distributions which conceives of sampling
distributions in terms of relative frequency distributions. Specifically, the
relative frequency distribution that would occur if samples of two balls
were repeatedly taken and the mean of each sample computed.

When we have a truly continuous distribution, it is not only impractical but
actually impossible to enumerate all possible outcomes. Moreover, in
continuous distributions, the probability of obtaining any single value is
zero. Therefore, as discussed in our introduction to Distributions, these
values are called probability densities rather than probabilities.

Sampling Distributions and Inferential Statistics

As we stated in the beginning of this chapter, sampling distributions are
important for inferential statistics. In the examples given so far, a
population was specified and the sampling distribution of the mean and the
range were determined. In practice, the process proceeds the other way: you
collect sample data and, from these data, you estimate parameters of the
sampling distribution. This knowledge of the sampling distribution can be
very useful. For example, knowing the degree to which means from
different samples would differ from each other and from the population
mean would give you a sense of how close your particular sample mean is
likely to be to the population mean. Fortunately, this information is directly
available from a sampling distribution: The most common measure of how
much sample means differ from each other is the standard deviation of the
sampling distribution of the mean. This standard deviation is called the

standard error of the mean. If all the sample means were very close to the
population mean, then the standard error of the mean would be small. On
the other hand, if the sample means varied considerably, then the standard
error of the mean would be large.

To be specific, assume your sample mean were 125 and you estimated that
the standard error of the mean were 5 (using a method shown in a later
section). If you had a normal distribution, then it would be likely that your
sample mean would be within 10 units of the population mean since most of
a normal distribution is within two standard deviations of the mean.

Keep in mind that all statistics have sampling distributions, not just the
mean. In later sections we will be discussing the sampling distribution of
the variance, the sampling distribution of the difference between means, and
the sampling distribution of Pearson's correlation, among others.

The Central Limit Theorem for Sample Means (Averages)

Suppose X is a random variable with a distribution that may be known or
unknown (it can be any distribution). Using a subscript that matches the
random variable, suppose:

° ajx = the mean of X
e boy = the standard deviation of X

If you draw random samples of size n, then as n increases, the random

variable X which consists of sample means, tends to be normally
distributed and

X~ N(x, )

The Central Limit Theorem for Sample Means says that if you keep
drawing larger and larger samples (like rolling 1, 2, 5, and, finally, 10 dice)
and calculating their means the sample means form their own normal
distribution (the sampling distribution). The normal distribution has the
Same mean as the original distribution and a variance that equals the
original variance divided by n, the sample size. n is the number of values
that are averaged together not the number of times the experiment is done.

To put it more formally, if you draw random samples of size n,the
distribution of the random variable X , which consists of sample means, is
called the sampling distribution of the mean. The sampling distribution of
the mean approaches a normal distribution as n, the sample size, increases.

The random variable X has a different z-score associated with it than the
random variable X. x is the value of X in one sample.
Equation:

wv PX

@

—

tx is both the average of X and of X.

Oy = a4 = standard deviation of X and is called the standard error of

the mean.

Example:

An unknown distribution has a mean of 90 and a standard deviation of 15.
Samples of size n = 25 are drawn randomly from the population.
Exercise:

Problem:

Find the probability that the sample mean is between 85 and 92.
Solution:

Let X = one value from the original unknown population. The
probability question asks you to find a probability for the sample

mean.

Let X = the mean of a sample of size 25. Since wx = 90, 0x = 15,
andn = 25;

. “15
then X N(90, 45.)

Find P(85 < x < 92) Draw a graph.

P(85 < a < 92) = 0.6997

The probability that the sample mean is between 85 and 92 is 0.6997.

P(8S < x < 92)

A |

85 90 92

TI-83 or 84: normalcdf (lower value, upper value, mean, standard
error of the mean)

The parameter list is abbreviated (lower value, upper value, J, Wad

normalcdf (85,92,90, = ) = 0.6997
Exercise:
Problem:

Find the value that is 2 standard deviations above the expected value
(it is 90) of the sample mean.

Solution:

To find the value that is 2 standard deviations above the expected
value 90, use the formula

value = wx + (#o0fSTDEVs) (+)

n

: eee
value = 90 + 2 FE 96

So, the value that is 2 standard deviations above the expected value is
96.

Example:

The length of time, in hours, it takes an "over 40" group of people to play
one soccer match is normally distributed with a mean of 2 hours and a
standard deviation of 0.5 hours. A sample of size n = 50 is drawn
randomly from the population.

Exercise:

Problem:

Find the probability that the sample mean is between 1.8 hours and
2.3 hours.

Solution:
Let X = the time, in hours, it takes to play one soccer match.

The probability question asks you to find a probability for the sample
mean time, in hours, it takes to play one soccer match.

Let X = the mean time, in hours, it takes to play one soccer match.

oie os a Sandi? — , then
X ~N( ; ) by the Central Limit Theorem for Means.

ix —2,0,% — 0.5, 77 — 50, and X-N(2, 45)
Findbe eS 253), Draw a graph.
GL 2a 2 AS) SO

pee eee
normalcdf(1.8,2.3,2, a5) — 0.9977

The probability that the mean time is between 1.8 hours and 2.3 hours
is

Glossary

Average
A number that describes the central tendency of the data. There are a
number of specialized averages, including the arithmetic mean,
weighted mean, median, mode, and geometric mean.

Central Limit Theorem
Given a random variable (RV) with known mean pu and known
standard deviation o. We are sampling with size n and we are
interested in two new RVs - the sample mean, X, and the sample sum,
3/X. If the size n of the sample is sufficiently large, then X ~

N(n, <) and XX ~ N (au, Jno). If the size n of the sample is

sufficiently large, then the distribution of the sample means and the
distribution of the sample sums will approximate a normal distribution
regardless of the shape of the population. The mean of the sample
means will equal the population mean and the mean of the sample
sums will equal n times the population mean. The standard deviation
of the distribution of the sample means, Ta is called the standard

error of the mean.

Normal Distribution
A continuous random variable (RV) with pdf

Ex) = rs e~(*-H)"/20° | where jy. is the mean of the distribution and

o is the standard deviation. Notation: X ~ N(y,o). If uw = 0 and
ao = 1, the RV is called the standard normal distribution.

Standard Error of the Mean

The standard deviation of the distribution of the sample means, —

ane

The Central Limit Theorem for Sums

Suppose X is a random variable with a distribution that may be known or
unknown (it can be any distribution) and suppose:

° ax =the mean of X
e boy = the standard deviation of X

If you draw random samples of size n, then as n increases, the random
variable 4X which consists of sums tends to be normally distributed and

UX ~ N(n- px,./n- ox)

The Central Limit Theorem for Sums says that if you keep drawing
larger and larger samples and taking their sums, the sums form their own
normal distribution (the sampling distribution) which approaches a normal
distribution as the sample size increases. The normal distribution has a
mean equal to the original mean multiplied by the sample size and a
standard deviation equal to the original standard deviation multiplied
by the square root of the sample size.

The random variable _X has the following z-score associated with it:

e a» x is one sum.
e bz = 22amex
J/n-ox

° an- wx = the mean of © X
e b./n- ox = standard deviation of X

Example:

An unknown distribution has a mean of 90 and a standard deviation of 15.
A sample of size 80 is drawn randomly from the population.

Exercise:

Problem:

¢ aFind the probability that the sum of the 80 values (or the total of
the 80 values) is more than 7500.
e bFind the sum that is 1.5 standard deviations above the mean of
the sums.
Solution:
Let X = one value from the original unknown population. The
probability question asks you to find a probability for the sum (or

total of) 80 values.

“X = the sum or total of 80 values. Since wx = 90, 7x = 15, and
n = 80, then

ee N(80 - 90, 80 - 15)

e mean of the sums = n- 4x = (80)(90) = 7200
e standard deviation of the sums = \/n- ox = V80- 15
e sum of 80 values = x = 7500

e aFind P(=x > 7500)

JE( Dee 7151010) UA

P > % > 7500

yx

7200 7500

normalcdf (lower value, upper value, mean of sums, stdev of
sums)

The parameter list is abbreviated (lower, upper, n - x, \/n- ox)

normalcdf(7500,1E99, 80 - 90, 80 - 15) = 0.0127
Reminder: 1E99 = 10°°. Press the EE key for E.

e bFind ix where z = 1.5:
Ux=n-px t+ z-./n-ox = (80)(90) + (1.5)(V/80) (15) = 7401.2

Glossary

Central Limit Theorem
Given a random variable (RV) with known mean p and known
standard deviation o. We are sampling with size n and we are
interested in two new RVs - the sample mean, X, and the sample sum,
3/X. If the size n of the sample is sufficiently large, then X ~

N(n, a and 3X ~ N (nu, Jno). If the size n of the sample is

sufficiently large, then the distribution of the sample means and the
distribution of the sample sums will approximate a normal distribution
regardless of the shape of the population. The mean of the sample
means will equal the population mean and the mean of the sample
sums will equal n times the population mean. The standard deviation

of the distribution of the sample means, —~, is called the standard

Jn ’

error of the mean.

Normal Distribution

A continuous random variable (RV) with pdf

f(x) = TE e~(=—-H)"/20° where js is the mean of the distribution and

o is the standard deviation. Notation: X ~ N(w,o). If u = 0 and
ao = 1, the RV is called the standard normal distribution.

Using the Central Limit Theorem

Central Limit Theorem: Using the Central Limit Theorem is part of the
collection col10555 written by Barbara Illowsky and Susan Dean. It covers
how and when to use the Central Limit Theorem and has contributions from
Roberta Bloom.

It is important for you to understand when to use the CLT. If you are being
asked to find the probability of the mean, use the CLT for the mean. If you
are being asked to find the probability of a sum or total, use the CLT for
sums. This also applies to percentiles for means and sums.

Note:If you are being asked to find the probability of an individual value,
do not use the CLT. Use the distribution of its random variable.

Examples of the Central Limit Theorem
Law of Large Numbers

The Law of Large Numbers says that if you take samples of larger and
larger size from any population, then the mean x of the sample tends to get
closer and closer to pw. From the Central Limit Theorem, we know that as n
gets larger and larger, the sample means follow a normal distribution. The
larger n gets, the smaller the standard deviation gets. (Remember that the

standard deviation for X is Va .) This means that the sample mean z must
be close to the population mean jz. We can say that yz is the value that the
sample means approach as n gets larger. The Central Limit Theorem

illustrates the Law of Large Numbers.

Central Limit Theorem for the Mean and Sum Examples

Example:

A study involving stress is done on a college campus among the students.
The stress scores follow a uniform distribution with the lowest stress
score equal to 1 and the highest equal to 5. Using a sample of 75 students,
find:

1. The probability that the mean stress score for the 75 students is less
than 2.

2. The 90th percentile for the mean stress score for the 75 students.

3. The probability that the total of the 75 stress scores is less than 200.

4. The 90th percentile for the total stress score for the 75 students.

Let X = one stress score.

Problems 1. and 2. ask you to find a probability or a percentile for a mean.
Problems 3 and 4 ask you to find a probability or a percentile for a total or
sum. The sample size, n, is equal to 75.

Since the individual stress scores follow a uniform distribution, X ~
U(1,5) where a = 1 and b = 5 (See Continuous Random Variables for the

year
ale 4f°. _ =

I= [te 4) GE oe = Lie

For iene 1. and 2., let X = the mean stress score for the 75 students.
Then,

Ee 115 =

x N(3, 115 ) where n = 75.

Exercise:
Problem: Find P(x < 2). Draw the graph.
Solution:
P(x <2)=0

The probability that the mean stress score is less than 2 is about 0.

p(x < 2)

normalcdf Woe 118 | = i
V75

Note:The smallest stress score is 1. Therefore, the smallest mean for
75 stress scores is 1.

Exercise:

Problem:

Find the 90th percentile for the mean of 75 stress scores. Draw a
graph.

Solution:
Let k = the 90th precentile.
Find k where P(x < k) = 0.90.

oS a

P(x < k)= 0.90

x

The 90th percentile for the mean of 75 scores is about 3.2. This tells
us that 90% of all the means of 75 stress scores are at most 3.2 and
10% are at least 3.2.

1 75
For problems c and d, let 3:X = the sum of the 75 stress scores. Then, 1X
. N|(75) (2), Nae 1.15 |

Exercise:

invNorm (.90, 3 LAs ) = 3.2

Problem: Find P(x < 200). Draw the graph.

Solution:
The mean of the sum of 75 stress scores is 75 - 3 = 225

The standard deviation of the sum of 75 stress scores is
V/75-1.15 = 9.96

P(=x < 200) =0

| = 00

200 225 ms

The probability that the total of 75 scores is less than 200 is about 0.

normalcdf (75, 200, 75 3,75 - 1.15) =.

Note: The smallest total of 75 stress scores is 75 since the smallest
single score is 1.

Exercise:

Problem:

Find the 90th percentile for the total of 75 stress scores. Draw a graph.
Solution:

Let k = the 90th percentile.

Find k where P(Xx < k) = 0.90.

k = 237.8

225 &k =

The 90th percentile for the sum of 75 scores is about 237.8. This tells
us that 90% of all the sums of 75 scores are no more than 237.8 and
10% are no less than 237.8.

invNorm (.90, 75 aoe Toe 1.15) ~ 237.8

Example:

Suppose that a market research analyst for a cell phone company conducts
a study of their customers who exceed the time allowance included on their
basic cell phone contract; the analyst finds that for those people who
exceed the time included in their basic contract, the excess time used
follows an exponential distribution with a mean of 22 minutes.

Consider a random sample of 80 customers who exceed the time allowance
included in their basic cell phone contract.

Let X = the excess time used by one INDIVIDUAL cell phone customer
who exceeds his contracted time allowance.

DS Exp(z;) From Chapter 5, we know that = 22 ando = 22.

Let X = the mean excess time used by a sample of n = 80 customers who
exceed their contracted time allowance.

xX ~ N(22, 22.) by the CLT for Sample Means
Exercise:

Problem:

Using the CLT to find Probability:

e aFind the probability that the mean excess time used by the 80
customers in the sample is longer than 20 minutes. This is asking
us to find P(a > 20) Draw the graph.

¢ b Suppose that one customer who exceeds the time limit for his
cell phone contract is randomly selected. Find the probability
that this individual customer's excess time is longer than 20
minutes. This is asking us to find P(a > 20)

e c Explain why the probabilities in (a) and (b) are different.

Solution:
Part a.

iehinvol: J24y = ZA0))

P(x > 20) = 0.7919 using normalcdf (20, 1E99, 22, 22.)

The probability is 0.7919 that the mean excess time used is more than
20 minutes, for a sample of 80 customers who exceed their contracted
time allowance.

p(x > 20)

|

20 22

Note:1E99 = 10°°and-1E99 = —10””. Press the
EE

key for E. Or just use 10499 instead of 1E99.

Part b.
Find P(x>20) . Remember to use the exponential distribution for an
individual: X~Exp(1/22).

P(X>20) = e\(-(1/22)*20) or e\(—.04545*20) = 0.4029
Part c. Explain why the probabilities in (a) and (b) are different.

¢ P(x > 20) = 0.4029 but P(x > 20) = 0.7919

e The probabilities are not equal because we use different
distributions to calculate the probability for individuals and for
means.

e When asked to find the probability of an individual value, use the
stated distribution of its random variable; do not use the CLT.
Use the CLT with the normal distribution when you are being
asked to find the probability for an mean.

Exercise:

Problem:

Using the CLT to find Percentiles:

Find the 95th percentile for the sample mean excess time for samples
of 80 customers who exceed their basic contract time allowances.
Draw a graph.

Solution:

Let k = the 95th percentile. Find k where P(x < k) = 0.95

a =, 22) _
k = 26.0 using invNorm(.95, 22, 22.) — 26.0

P(x <k)= 0.95

x
22 k

The 95th percentile for the sample mean excess time used is about
26.0 minutes for random samples of 80 customers who exceed their
contractual allowed time.

95% of such samples would have means under 26 minutes; only 5%
of such samples would have means above 26 minutes.

Note:(HISTORICAL): Normal Approximation to the Binomial

Historically, being able to compute binomial probabilities was one of the
most important applications of the Central Limit Theorem. Binomial
probabilities were displayed in a table in a book with a small value for n
(say, 20). To calculate the probabilities with large values of n, you had to
use the binomial formula which could be very complicated. Using the
Normal Approximation to the Binomial simplified the process. To
compute the Normal Approximation to the Binomial, take a simple random
sample from a population. You must meet the conditions for a binomial
distribution:

e there are a certain number n of independent trials
e the outcomes of any trial are success or failure

¢ each trial has the same probability of a success p

Recall that if X is the binomial random variable, then X~B(n, p). The
shape of the binomial distribution needs to be similar to the shape of the
normal distribution. To ensure this, the quantities np and nq must both be
greater than five (np > 5 and nq > 5; the approximation is better if they
are both greater than or equal to 10). Then the binomial can be
approximated by the normal distribution with mean yz = np and standard
deviation a = ,/npq. Remember that q = 1 — p. In order to get the best
approximation, add 0.5 to x or subtract 0.5 from x (use x + 0.5 or x — 0.5
). The number 0.5 is called the continuity correction factor.

Example:

Suppose in a local Kindergarten through 12th grade (K - 12) school
district, 53 percent of the population favor a charter school for grades K -
5. A simple random sample of 300 is surveyed.

1. Find the probability that at least 150 favor a charter school.

2. Find the probability that at most 160 favor a charter school.

3. Find the probability that more than 155 favor a charter school.
4. Find the probability that less than 147 favor a charter school.
5. Find the probability that exactly 175 favor a charter school.

Let X = the number that favor a charter school for grades K - 5. X~
B(n, p) where n =300 and p = 0.53. Since np > 5 and nq > 5, use the
normal approximation to the binomial. The formulas for the mean and
standard deviation are 4. = np and 0 = ,/npq. The mean is 159 and the
standard deviation is 8.6447. The random variable for the normal
distribution is Y. Y~N(159, 8.6447). See The Normal Distribution for
help with calculator instructions.

For Problem 1., you include 150 so P(x > 150) has normal
approximation P(Y > 149.5) =0.8641.

normalcdf (149.5, 10°99, 159, 8.6447) = 0.8641.

For Problem 2., you include 160 so P(x < 160) has normal
approximation P(Y < 160.5) =0.5689.

normalcdf (0, 160.5, 159, 8.6447) = 0.5689
For Problem 3., you exclude 155 so P(x > 155) has normal
approximation P(y > 155.5) =0.6572.
normalcdf (155.5, 10°99, 159, 8.6447) = 0.6572
For Problem 4., you exclude 147 so P(x < 147) has normal
approximation P(Y < 146.5) =0.0741.
normalcdf (0, 146.5, 159, 8.6447) = 0.0741
For Problem 5., P(a=175) has normal approximation
VEN els ee 11) OO Fos},
normalcdf (174.5, 175.5, 159, 8.6447) = 0.0083
Because of calculators and computer software that easily let you
calculate binomial probabilities for large values of n, it is not necessary to
use the the Normal Approximation to the Binomial provided you have
access to these technology tools. Most school labs have Microsoft Excel,
an example of computer software that calculates binomial probabilities.
Many students have access to the TI-83 or 84 series calculators and they
easily calculate probabilities for the binomial. In an Internet browser, if
you type in "binomial probability distribution calculation," you can find at
least one online calculator for the binomial.
For Example 3, the probabilities are calculated using the binomial (n=300
and p=0.53) below. Compare the binomial and normal distribution
answers. See Discrete Random Variables for help with calculator
instructions for the binomial.
P(x > 150):1 - binomialcdf (300, 0.53, 149) =0.8641

P(x < 160): binomialcdf (300, 0.53, 160) =0.5684
Pe t55)1 - binomialcdT eno wss 5s) 06576
P(a < 147): binomialcdf (300, 0.53, 146) =0.0742
P(a=175): (You use the binomial pdf.) binomialpdf
(175,,0'52, 146) 070083

**Contributions made to Example 2 by Roberta Bloom

Glossary

Average
A number that describes the central tendency of the data. There are a
number of specialized averages, including the arithmetic mean,
weighted mean, median, mode, and geometric mean.

Central Limit Theorem
Given a random variable (RV) with known mean p and known
standard deviation o. We are sampling with size n and we are
interested in two new RVs - the sample mean, X, and the sample sum,
»/X. If the size n of the sample is sufficiently large, then X ~

N(n, =| and &.X ~ N(np, no). If the size n of the sample is

sufficiently large, then the distribution of the sample means and the
distribution of the sample sums will approximate a normal distribution
regardless of the shape of the population. The mean of the sample
means will equal the population mean and the mean of the sample
sums will equal n times the population mean. The standard deviation
of the distribution of the sample means, aia is called the standard

error of the mean.

Exponential Distribution
A continuous random variable (RV) that appears when we are
interested in the intervals of time between some random events, for
example, the length of time between emergency arrivals at a hospital.
Notation: X~Exp(m). The mean is ps = —- and the standard deviation

is 0 = —. The probability density function is f(x) = me ™, x > 0
and the cumulative distribution function is P(X < «) =1—e"™.

Mean
A number that measures the central tendency. A common name for
mean is 'average.' The term 'mean' is a shortened form of ‘arithmetic

mean.’ By definition, the mean for a sample (denoted by 2) is
Sum of all values in the sample dth f lati
Number of values in the sample ’ an See Or popiawon

: __ Sum of all values in the population
(denoted by Lt) 1S & = Number of values in the population °

C=

Uniform Distribution

A continuous random variable (RV) that has equally likely outcomes
over the domain, a < x < b. Often referred as the Rectangular
distribution because the graph of the pdf has the form of a rectangle.

Notation: X~U(a,b). The mean is p = ath and the standard deviation
iso = = Y The probability density function is f(a) = - for

a<2a<bora<z< b. The cumulative distribution is

P(X <a) = $2,

Summary of Formulas
Formula
Central Limit Theorem for Sample Means

X~N(ux, ) The Mean (x): [Lx
Formula

Central Limit Theorem for Sample Means Z-Score and Standard Error of
the Mean

z= (& ) Standard Error of the Mean (Standard Deviation
Vn
(x):

Formula
Central Limit Theorem for Sums

uX~ N|(n)-ux,/n-ox| Mean for Sums (DX): n- ux
Formula
Central Limit Theorem for Sums Z-Score and Standard Deviation for Sums

Ux—n- px

VMox

= Standard Deviation for Sums (»X): J/n-ox

Practice: The Central Limit Theorem

Student Learning Outcomes

e The student will calculate probabilities using the Central Limit
Theorem.

Given

Yoonie is a personnel manager in a large corporation. Each month she must
review 16 of the employees. From past experience, she has found that the
reviews take her approximately 4 hours each to do with a population
standard deviation of 1.2 hours. Let X be the random variable representing
the time it takes her to complete one review. Assume X is normally
distributed. Let X be the random variable representing the mean time to
complete the 16 reviews. Let 4X be the total time it takes Yoonie to
complete all of the month’s reviews. Assume that the 16 reviews represent a
random set of reviews.

Distribution
Complete the distributions.

le. ee
Die i
Be eee

Graphing Probability
For each problem below:
e a Sketch the graph. Label and scale the horizontal axis. Shade the
region corresponding to the probability.

e b Calculate the value.

Exercise:

Problem:

Find the probability that one review will take Yoonie from 3.5 to 4.25
hours.

ea

: ye

Solution:
e b3.5, 4.25, 0.2441
Exercise:
Problem:

Find the probability that the mean of a month’s reviews will take
Yoonie from 3.5 to 4.25 hrs.

ea

4 |

© b P( =

Solution:
e b 0.7499
Exercise:
Problem:

Find the 95th percentile for the mean time to complete one month’s
reviews.

ea

i |

e bThe 95th Percentile=

Solution:
e b 4.49 hours
Exercise:
Problem:

Find the probability that the sum of the month’s reviews takes Yoonie
from 60 to 65 hours.

ea

¢ b The Probability=

Solution:
e b 0.3802

Exercise:

Problem: Find the 95th percentile for the sum of the month’s reviews.

ea

e b The 95th percentile=

Solution:

¢ b 71.90

Discussion Question

Exercise:

Problem: What causes the probabilities in [link] and [link] to differ?

Homework

The Central Limit Theorem: Homework is part of the collection col10555
written by Barbara Illowsky and Susan Dean.

Exercise:

Problem:

X ~ N(60,9). Suppose that you form random samples of 25 from this

distribution. Let X be the random variable of averages. Let »X be the
random variable of sums. For c - f, sketch the graph, shade the region,

label and scale the horizontal axis for X,, and find the probability.

e a Sketch the distributions of X and X on the same graph.
ebx~

e c P(x < 60) =

e d Find the 30th percentile for the mean.

e e P(56 < x < 62) =

‘fPUs8. <2 < 538) =

° g Ux~

e h Find the minimum value for the upper quartile for the sum.
i P(1400' = Se < 1550) =

Solution:

¢ b Xbar~N(60,—_)

, /25
c0.5000
d59.06
e0.8536
£0.1333
h1530.35
10.8536

Exercise:

Problem:

Determine which of the following are true and which are false. Then,
in complete sentences, justify your answers.

e a When the sample size is large, the mean of X is approximately
equal to the mean of X.

¢ b When the sample size is large, X is approximately normally
distributed.

e c When the sample size is large, the standard deviation of X is
approximately the same as the standard deviation of X.

Exercise:

Problem:

The percent of fat calories that a person in America consumes each
day is normally distributed with a mean of about 36 and a standard
deviation of about 10. Suppose that 16 individuals are randomly
chosen.

Let X =average percent of fat calories.

eaX~ ( :

¢ b For the group of 16, find the probability that the average
percent of fat calories consumed is more than 5. Graph the
situation and shade in the area to be determined.

e c Find the first quartile for the average percent of fat calories.

Solution:
f _10_
a N(36, ag)
e bl
e € 34.31

Exercise:

Problem:

Previously, De Anza statistics students estimated that the amount of
change daytime statistics students carry is exponentially distributed
with a mean of $0.88. Suppose that we randomly pick 25 daytime
Statistics students.

a In words, X =

b X~

c In words, X =

dX~_____( )

e Find the probability that an individual had between $0.80 and
$1.00. Graph the situation and shade in the area to be determined.
f Find the probability that the average of the 25 students was
between $0.80 and $1.00. Graph the situation and shade in the
area to be determined.

g Explain the why there is a difference in (e) and (f).

’

Exercise:

Problem:

Suppose that the distance of fly balls hit to the outfield (in baseball) is
normally distributed with a mean of 250 feet and a standard deviation
of 50 feet. We randomly sample 49 fly balls.

a If X = average distance in feet for 49 fly balls, then X ~

b What is the probability that the 49 balls traveled an average of

less than 240 feet? Sketch the graph. Scale the horizontal axis for
X. Shade the region corresponding to the probability. Find the
probability.

c Find the 80th percentile of the distribution of the average of 49
fly balls.

Solution:

: 50
a N(250,)

e b 0.0808
e ¢c256.01 feet

Exercise:

Problem:

Suppose that the weight of open boxes of cereal in a home with
children is uniformly distributed from 2 to 6 pounds. We randomly
survey 64 homes with children.

e alin words, X =

b X~

C Ux =

do x=

e In words, XX =

f X~

e g Find the probability that the total weight of open boxes is less
than 250 pounds.

e h Find the 35th percentile for the total weight of open boxes of
cereal.

Exercise:

Problem:

Suppose that the duration of a particular type of criminal trial is known
to have a mean of 21 days and a standard deviation of 7 days. We
randomly sample 9 trials.

e alnwords, 1) X =

© bxyX~

e c Find the probability that the total length of the 9 trials is at least
225 days.

e d90 percent of the total of 9 of these types of trials will last at
least how long?

Solution:

e aThe total length of time for 9 criminal trials
¢ b N(189,21)

e ¢c 0.0432

e d 162.09

Exercise:

Problem:

According to the Internal Revenue Service, the average length of time
for an individual to complete (record keep, learn, prepare, copy,
assemble and send) IRS Form 1040 is 10.53 hours (without any
attached schedules). The distribution is unknown. Let us assume that
the standard deviation is 2 hours. Suppose we randomly sample 36
taxpayers.

e aln words, X =

e b In words, X =

ecxX~

e d Would you be surprised if the 36 taxpayers finished their Form
1040s in an average of more than 12 hours? Explain why or why
not in complete sentences.

e e Would you be surprised if one taxpayer finished his Form 1040
in more than 12 hours? In a complete sentence, explain why.

Exercise:
Problem:
Suppose that a category of world class runners are known to run a
marathon (26 miles) in an average of 145 minutes with a standard

deviation of 14 minutes. Consider 49 of the races.

Let X = the average of the 49 races.

eaX~-

e b Find the probability that the runner will average between 142
and 146 minutes in these 49 marathons.

e c Find the 80th percentile for the average of these 49 marathons.

e d Find the median of the average running times.

Solution:
14
e b 0.6247
e c 146.68

e d 145 minutes

Exercise:

Problem:

The attention span of a two year-old is exponentially distributed with a
mean of about 8 minutes. Suppose we randomly survey 60 two year-
olds.

e alnwords, X =

° bxX-~

e cIn words, X =

ed X~-

e e Before doing any calculations, which do you think will be
higher? Explain why.

© ithe probability that an individual attention span is less than
10 minutes; or

© ii the probability that the average attention span for the 60
children is less than 10 minutes? Why?

e f Calculate the probabilities in part (e).
e g Explain why the distribution for X is not exponential.

Exercise:

Problem:

Suppose that the length of research papers is uniformly distributed
from 10 to 25 pages. We survey a class in which 55 research papers
were turned in to a professor. The 55 research papers are considered a
random collection of all papers. We are interested in the average length
of the research papers.

a In words, X =

b X~

C Ux =

do x=

e In words, X =

f X -

g In words, 7X =

h )X ~

i Without doing any calculations, do you think that it’s likely that
the professor will need to read a total of more than 1050 pages?
Why?

j Calculate the probability that the professor will need to read a
total of more than 1050 pages.

k Why is it so unlikely that the average length of the papers will
be less than 12 pages?

Solution:

b U(10,25)
c17.5

ed 1/2 = 4.3301

£N(17.5,0.5839)
hN(962.5,32.11)
j0.0032

Exercise:

Problem:

The length of songs in a collector’s CD collection is uniformly
distributed from 2 to 3.5 minutes. Suppose we randomly pick 5 CDs
from the collection. There is a total of 43 songs on the 5 CDs.

e alnwords, X =

° b X-

e clin words, X =

dx

e e Find the first quartile for the average song length.

e f The IQR (interquartile range) for the average song length is
from to

Exercise:

Problem:

Salaries for teachers in a particular elementary school district are
normally distributed with a mean of $44,000 and a standard deviation
of $6500. We randomly survey 10 teachers from that district.

e alnwords, X =

¢ bIn words, X =

ecX-

e din words, :X =

ee diX~

e f Find the probability that the teachers earn a total of over
$400,000.

e g Find the 90th percentile for an individual teacher’s salary.

e h Find the 90th percentile for the average teachers’ salary.

e ilf we surveyed 70 teachers instead of 10, graphically, how
would that change the distribution for X ?

e j If each of the 70 teachers received a $3000 raise, graphically,
how would that change the distribution for X ?

Solution:

2 c N(44,000, “*)

¢ e N(440,000,(/10)(6500))
° £0.9742

¢ g $52,330

¢ h $46,634

Exercise:

Problem:

The distribution of income in some Third World countries is
considered wedge shaped (many very poor people, very few middle
income people, and few to many wealthy people). Suppose we pick a
country with a wedge distribution. Let the average salary be $2000 per
year with a standard deviation of $8000. We randomly survey 1000
residents of that country.

e alnwords, X =

e bIn words, X =

ecX-

¢ d How is it possible for the standard deviation to be greater than
the average?

e Why is it more likely that the average of the 1000 residents will
be from $2000 to $2100 than from $2100 to $2200?

Exercise:
Problem:
The average length of a maternity stay in a U.S. hospital is said to be

2.4 days with a standard deviation of 0.9 days. We randomly survey 80
women who recently bore children in a U.S. hospital.

e alnwords, X =
e b In words, X =

c X ~

d In words, 3X =

e LX ~

f Is it likely that an individual stayed more than 5 days in the

hospital? Why or why not?

e g Is it likely that the average stay for the 80 women was more
than 5 days? Why or why not?

e h Which is more likely:

© jan individual stayed more than 5 days; or
o jithe average stay of 80 women was more than 5 days?

e i lf we were to sum up the women’s stays, is it likely that,
collectively they spent more than a year in the hospital? Why or
why not?

Solution:
- 0.9
c N(2.4, Ja
e eNV(192,8.05)
e hindividual

Exercise:

Problem:

In 1940 the average size of a U.S. farm was 174 acres. Let’s say that
the standard deviation was 55 acres. Suppose we randomly survey 38
farmers from 1940. (Source: U.S. Dept. of Agriculture)

e aln words, X =

¢ bIn words, X =

ec xX~

e d The IQR for X is from acres to acres.

Exercise:

Problem:

The stock closing prices of 35 U.S. semiconductor manufacturers are
given below. (Source: Wall Street Journal)

8.625 30.25 27.625 46.75 32.875 18.25 5 0.125 2.9375 6.875 28.25
24,25.21.1.5 30.25.71 43.5 49.25:2:5629'31 16.5.9:5 18.918: 9:10:5
16,625°1.25°18:12.875 7.12875 2,875 '60.29:29.25

a In words, X =
b

°oiz=
foley | eee
° in =

c Construct a histogram of the distribution of the averages. Start
at x = —0.0005. Make bar widths of 10.

d In words, describe the distribution of stock prices.

e Randomly average 5 stock prices together. (Use a random
number generator.) Continue averaging 5 pieces together until
you have 10 averages. List those 10 averages.

f Use the 10 averages from (e) to calculate:

10) iz-=
° 1S, =

g Construct a histogram of the distribution of the averages. Start
at = —0.0005. Make bar widths of 10.

h Does this histogram look like the graph in (c)?

i In 1 - 2 complete sentences, explain why the graphs either look
the same or look different?

j Based upon the theory of the Central Limit Theorem, X ~

Solution:

b$20.71; $17.31; 35

¢ d Exponential distribution, X ~ Exp(1/20.71)
e £ $20.71; $11.14

e j N(20.71, at)

Exercise:

Problem:

Use the Initial Public Offering data (see “Table of Contents) to do this
problem.

e alnwords, X =
« b

(e) iwx =
ie) liox =
o in =

e c Construct a histogram of the distribution. Start at = —0.50.
Make bar widths of $5.

e dIn words, describe the distribution of stock prices.

e e Randomly average 5 stock prices together. (Use a random
number generator.) Continue averaging 5 pieces together until
you have 15 averages. List those 15 averages.

e f Use the 15 averages from (e) to calculate the following:

°oiz=
°o is, =

¢ g Construct a histogram of the distribution of the averages. Start
at x = —0.50. Make bar widths of $5.

e h Does this histogram look like the graph in (c)? Explain any
differences.

e iln 1-2 complete sentences, explain why the graphs either look
the same or look different?

e j Based upon the theory of the Central Limit Theorem, X ~

Try these multiple choice questions (Exercises19 - 23).

The next two questions refer to the following information: The time to
wait for a particular rural bus is distributed uniformly from 0 to 75 minutes.
100 riders are randomly sampled to learn how long they waited.

Exercise:

Problem:

The 90th percentile sample average wait time (in minutes) for a sample
of 100 riders is:

e A 315.0
¢ B40.3
e C 38.5
e D 65.2

Solution:

B
Exercise:

Problem:

Would you be surprised, based upon numerical calculations, if the
sample average wait time (in minutes) for 100 riders was less than 30
minutes?

e A Yes
¢ BNo
e C There is not enough information.

Solution:

A

Exercise:

Problem:

Which of the following is NOT TRUE about the distribution for
averages?

e A The mean, median and mode are equal
e B The area under the curve is one

e C The curve never touches the x-axis

e D The curve is skewed to the right

Solution:

D

The next three questions refer to the following information: The cost of
unleaded gasoline in the Bay Area once followed an unknown distribution
with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations
from the Bay Area are randomly chosen. We are interested in the average
cost of gasoline for the 16 gas stations.

Exercise:

Problem:
The distribution to use for the average cost of gasoline for the 16 gas
Stations is

e AX ~ N(4.59, 0.10)

Rx e N(4.59, 2.0.)

V16
© CX ~N(4.59, 2)
°DX~N

(4.59, “5
(

4.59, se)

Solution:

B

Exercise:
Problem:

What is the probability that the average price for 16 gas stations is over
$4.69?

e A Almost zero
¢ B0.1587

¢ € 0.0943

e D Unknown

Solution:

A
Exercise:
Problem:

Find the probability that the average price for 30 gas stations is less
than $4.55.

e A0.6554
¢ B0.3446
e €0.0142
e D0.9858
e E0

Solution:

C

Exercise:

Problem:

For the Charter School Problem (Example 6) in Central Limit
Theorem: Using the Central Limit Theorem, calculate the following
using the normal approximation to the binomial.

e A Find the probability that less than 100 favor a charter school for
grades K - 5.

¢ B Find the probability that 170 or more favor a charter school for
grades K - 5.

e C Find the probability that no more than 140 favor a charter
school for grades K - 5.

¢ D Find the probability that there are fewer than 130 that favor a
charter school for grades K - 5.

e E Find the probability that exactly 150 favor a charter school for
grades K - 5.

If you either have access to an appropriate calculator or computer
software, try calculating these probabilities using the technology. Try
also using the suggestion that is at the bottom of Central Limit
Theorem: Using the Central Limit Theorem for finding a website
that calculates binomial probabilities.

Solution:

e C 0.0162
e E 0.0268

Exercise:

Problem:

Four friends, Janice, Barbara, Kathy and Roberta, decided to carpool
together to get to school. Each day the driver would be chosen by
randomly selecting one of the four names. They carpool to school for
96 days. Use the normal approximation to the binomial to calculate the
following probabilities. Round the standard deviation to 4 decimal
places.

e A Find the probability that Janice is the driver at most 20 days.

e B Find the probability that Roberta is the driver more than 16
days.

e C Find the probability that Barbara drives exactly 24 of those 96
days.

If you either have access to an appropriate calculator or computer
software, try calculating these probabilities using the technology. Try
also using the suggestion that is at the bottom of Central Limit
Theorem: Using the Central Limit Theorem for finding a website
that calculates binomial probabilities.

Solution:

¢ A 0.2047
e B0.9615
e C 0.0938

**Fxercise 24 contributed by Roberta Bloom

Review

Central Limit Theorem: Review is part of the collection col10555 written
by Barbara Illowsky and Susan Dean. The module consists of review
exercises.

The next three questions refer to the following information: Richard’s
Furniture Company delivers furniture from 10 A.M. to 2 P.M. continuously
and uniformly. We are interested in how long (in hours) past the 10 A.M.
Start time that individuals wait for their delivery.

Exercise:

Problem: X ~

e AU(0,4)
* BU(10,2)
e C Exp(2)
D N(2,1)

Solution:

A

Exercise:

Problem: The average wait time is:

e Ai hour
e B2 hour
e C 2.5 hour
¢ D 4 hour

Solution:

B

Exercise:

Problem:

Suppose that it is now past noon on a delivery day. The probability that
a person must wait at least 1 5 more hours is:

>

ee
eo IY a BR > >
o0| cos] cot]

Solution:

A

Exercise:

Problem: Given: X~Exp(+).

e aFind P(z > 1)
e b Calculate the minimum value for the upper quartile.

¢ c Find P(z = $)

Solution:

e a0.7165
e b4.16
e c0

Exercise:
Problem:
e 40% of full-time students took 4 years to graduate

e 30% of full-time students took 5 years to graduate
e 20% of full-time students took 6 years to graduate

e 10% of full-time students took 7 years to graduate

The expected time for full-time students to graduate is:

e A4 years
e B 4.5 years
e C5 years
e D5.5 years

Solution:

C
Exercise:

Problem:

Which of the following distributions is described by the following
example?

Many people can run a short distance of under 2 miles, but as the
distance increases, fewer people can run that far.

e A Binomial

e B Uniform

e C Exponential
e D Normal

Solution:

G

Exercise:

Problem:

The length of time to brush one’s teeth is generally thought to be
exponentially distributed with a mean of & minutes. Find the
probability that a randomly selected person brushes his/her teeth less
than a minutes.

e AO.5
3

e Bs

¢ C0.43

¢ D0.63

Solution:

D
Exercise:

Problem:
Which distribution accurately describes the following situation?

The chance that a teenage boy regularly gives his mother a kiss
goodnight (and he should!!) is about 20%. Fourteen teenage boys are
randomly surveyed.

X =the number of teenage boys that regularly give their mother a kiss
goodnight

¢ A B(14,0.20)
© BP(2.8)
© C N(2.8,2.24)

° DExp( eT )

Solution:

A
Exercise:

Problem:
Which distribution accurately describes the following situation?

A 2008 report on technology use states that approximately 20 percent
of U.S. households have never sent an e-mail. (source:

http://www. webguild.org/2008/05/20-percent-of-americans-have-
never-used-email.php) Suppose that we select a random sample of
fourteen U.S. households .

X =the number of households in a 2008 sample of 14 households that
have never sent an email

¢ A B(14,0.20)
© BP(2.8)
© C N(2.8,2.24)

¢ D Exp( 750 )

Solution:

A

**Exercise 9 contributed by Roberta Bloom

Introduction

Student Learning Outcomes
By the end of this chapter, the student should be able to:

¢ Calculate and interpret confidence intervals for one population mean
and one population proportion.

e Interpret the student-t probability distribution as the sample size
changes.

e Discriminate between problems applying the normal and the student-t
distributions.

Introduction

Suppose you are trying to determine the mean rent of a two-bedroom
apartment in your town. You might look in the classified section of the
newspaper, write down several rents listed, and average them together. You
would have obtained a point estimate of the true mean. If you are trying to
determine the percent of times you make a basket when shooting a
basketball, you might count the number of shots you make and divide that
by the number of shots you attempted. In this case, you would have
obtained a point estimate for the true proportion.

We use sample data to make generalizations about an unknown population.
This part of statistics is called inferential statistics. The sample data help
us to make an estimate of a population parameter. We realize that the
point estimate is most likely not the exact value of the population
parameter, but close to it. After calculating point estimates, we construct
confidence intervals in which we believe the parameter lies.

In this chapter, you will learn to construct and interpret confidence
intervals. You will also learn a new distribution, the Student's-t, and how it
is used with these intervals. Throughout the chapter, it is important to keep
in mind that the confidence interval is a random variable. It is the parameter
that is fixed.

If you worked in the marketing department of an entertainment company,
you might be interested in the mean number of compact discs (CD's) a
consumer buys per month. If so, you could conduct a survey and calculate
the sample mean, Z, and the sample standard deviation, s. You would use
to estimate the population mean and s to estimate the population standard
deviation. The sample mean, 2%, is the point estimate for the population
mean, p. The sample standard deviation, s, is the point estimate for the
population standard deviation, o.

Each of #% and s is also called a statistic.

A confidence interval is another type of estimate but, instead of being just
one number, it is an interval of numbers. The interval of numbers is a range
of values calculated from a given set of sample data. The confidence
interval is likely to include an unknown population parameter.

Suppose for the CD example we do not know the population mean p but we
do know that the population standard deviation is o = 1 and our sample
size is 100. Then by the Central Limit Theorem, the standard deviation for
the sample mean is

Co 1
Vn —_ “V100 — 0.1.
The Empirical Rule, which applies to bell-shaped distributions, says that in
approximately 95% of the samples, the sample mean, 2%, will be within two
standard deviations of the population mean jz. For our CD example, two
standard deviations is (2)(0.1) = 0.2. The sample mean 7 is likely to be
within 0.2 units of p.

Because % is within 0.2 units of 44, which is unknown, then yp is likely to be
within 0.2 units of Z in 95% of the samples. The population mean pu is
contained in an interval whose lower number is calculated by taking the
sample mean and subtracting two standard deviations ((2)(0.1)) and whose
upper number is calculated by taking the sample mean and adding two
standard deviations. In other words, pz is between x — 0.2 and z + 0.2 in
95% of all the samples.

For the CD example, suppose that a sample produced a sample mean x = 2
. Then the unknown population mean yp is between

£=0,2=2=02 = Léande+0.2= 2+ 0.2 = 2.2

We say that we are 95% confident that the unknown population mean
number of CDs is between 1.8 and 2.2. The 95% confidence interval is
(1.8, 2.2).

The 95% confidence interval implies two possibilities. Either the interval
(1.8, 2.2) contains the true mean yp or our sample produced an Z that is not
within 0.2 units of the true mean pw. The second possibility happens for only
5% of all the samples (100% - 95%).

Remember that a confidence interval is created for an unknown population
parameter like the population mean, jz. Confidence intervals for some
parameters have the form

(point estimate - margin of error, point estimate + margin of error)

The margin of error depends on the confidence level or percentage of
confidence.

When you read newspapers and journals, some reports will use the phrase
"margin of error." Other reports will not use that phrase, but include a
confidence interval as the point estimate + or - the margin of error. These
are two ways of expressing the same concept.

Note: Although the text only covers symmetric confidence intervals, there
are non-symmetric confidence intervals (for example, a confidence interval
for the standard deviation).

Optional Collaborative Classroom Activity

Have your instructor record the number of meals each student in your class
eats out in a week. Assume that the standard deviation is known to be 3
meals. Construct an approximate 95% confidence interval for the true mean
number of meals students eat out each week.

1. Calculate the sample mean.
2.0 = 3 and n = the number of students surveyed.

| ape
3. Construct the interval (z 2 Fare +2 Fi )
We say we are approximately 95% confident that the true average number
of meals that students eat out in a week is between and

Glossary

Confidence Interval (CI)
An interval estimate for an unknown population parameter. This
depends on:

e The desired confidence level.

e Information that is known about the distribution (for example,
known standard deviation).

e The sample and its size.

Inferential Statistics
Also called statistical inference or inductive statistics. This facet of
Statistics deals with estimating a population parameter based on a
sample statistic. For example, if 4 out of the 100 calculators sampled
are defective we might infer that 4 percent of the production is
defective.

Parameter
A numerical characteristic of the population.

Point Estimate
A single number computed from a sample and used to estimate a
population parameter.

Confidence Interval, Single Population Mean, Population Standard
Deviation Known, Normal

Confidence Intervals: Confidence Interval, Single Population Mean,
Population Standard Deviation Known, Normal is part of the collection
col10555 written by Barbara Illowsky and Susan Dean with contributions
from Roberta Bloom.

Calculating the Confidence Interval

To construct a confidence interval for a single unknown population mean pu
, where the population standard deviation is known, we need z as an
estimate for jz and we need the margin of error. Here, the margin of error is
called the error bound for a population mean (abbreviated EBM). The
sample mean z is the point estimate of the unknown population mean pu
The confidence interval estimate will have the form:

e (point estimate - error bound, point estimate + error bound) or, in
symbols,(z — EBM, z + EBM)

The margin of error depends on the confidence level (abbreviated CL). The
confidence level is often considered the probability that the calculated
confidence interval estimate will contain the true population parameter.
However, it is more accurate to state that the confidence level is the percent
of confidence intervals that contain the true population parameter when
repeated samples are taken. Most often, it is the choice of the person
constructing the confidence interval to choose a confidence level of 90% or
higher because that person wants to be reasonably certain of his or her
conclusions.

There is another probability called alpha (@). @ is related to the confidence
level CL. q@ is the probability that the interval does not contain the unknown
population parameter.

Mathematically, a + CL = 1.

Example:

e Suppose we have collected data from a sample. We know the sample
mean but we do not know the mean for the entire population.
e The sample mean is 7 and the error bound for the mean is 2.5.

2 = 7 andih Bh = 25.

The confidence interval is (7 — 2.5, 7 + 2.5); calculating the values gives
(AL D5).

If the confidence level (CL) is 95%, then we say that "We estimate with
95% confidence that the true value of the population mean is between 4.5
and 9.5."

A confidence interval for a population mean with a known standard
deviation is based on the fact that the sample means follow an
approximately normal distribution. Suppose that our sample has a mean of
x = 10 and we have constructed the 90% confidence interval (5, 15) where
EBM = 5.

To get a 90% confidence interval, we must include the central 90% of the
probability of the normal distribution. If we include the central 90%, we
leave out a total of a = 10% in both tails, or 5% in each tail, of the normal
distribution.

Confidence Level (CL) = 0.90

x= 10
EBM = 5
x = —— = 5
P 10 ” x+EBM = 15

pt is believed to be in the interval (5, 15) with 90% confidence.

To capture the central 90%, we must go out 1.645 "standard deviations" on
either side of the calculated sample mean. 1.645 is the z-score from a

Standard Normal probability distribution that puts an area of 0.90 in the
center, an area of 0.05 in the far left tail, and an area of 0.05 in the far right
tail.

It is important that the "standard deviation" used must be appropriate for the
parameter we are estimating. So in this section, we need to use the standard
deviation that applies to sample means, which is ae Ta is commonly

called the "standard error of the mean" in order to clearly distinguish the
standard deviation for a mean from the population standard deviation o.

In summary, as a result of the Central Limit Theorem:

e X is normally distributed, that is, X ~ N(x Xx, Fir )

e When the population standard deviation o is known, we use a
Normal distribution to calculate the error bound.

Calculating the Confidence Interval:

To construct a confidence interval estimate for an unknown population
mean, we need data from a random sample. The steps to construct and
interpret the confidence interval are:

¢ Calculate the sample mean x from the sample data. Remember, in this
section, we already know the population standard deviation o.

e Find the Z-score that corresponds to the confidence level.

e Calculate the error bound EBM

e Construct the confidence interval

e Write a sentence that interprets the estimate in the context of the
situation in the problem. (Explain what the confidence interval means,
in the words of the problem.)

We will first examine each step in more detail, and then illustrate the
process with some examples.

Finding z for the stated Confidence Level

When we know the population standard deviation o, we use a standard
normal distribution to calculate the error bound EBM and construct the
confidence interval. We need to find the value of z that puts an area equal to

the confidence level (in decimal form) in the middle of the standard normal
distribution Z~N(0,1).

The confidence level, CL, is the area in the middle of the standard normal
distribution. CL = 1 — a. So qa is the area that is split equally between the
two tails. Each of the tails contains an area equal to > .

The z-score that has an area to the right of 5 is denoted by zs

For example, when CL = 0.95 then a = 0.05 and > = 0.025 ; we write
£2 = £025

The area to the right of 2.995 is 0.025 and the area to the left of z.95 is 1-
0.025 = 0.975

Za = 29.925 = 1.96, using a calculator, computer or a Standard Normal
probability table.

Using the TI83, TI83+ or TI84+ calculator: invNorm(0.975, 0,1) = 1.96

CALCULATOR NOTE: Remember to use area to the LEFT of za ;in this

chapter the last two inputs in the invNorm command are 0,1 because you
are using a Standard Normal Distribution Z~N(0,1)

EBM: Error Bound
The error bound formula for an unknown population mean pz when the
population standard deviation o is known is

° EBM = zg - =

Constructing the Confidence Interval

e The confidence interval estimate has the format
(2 — EBM,z + EBM).

The graph gives a picture of the entire situation.

Ch-F > 5 =Ch--a= 1.

o. a
ry CL=l-a 9

x-EBM X x+EBM

Writing the Interpretation

The interpretation should clearly state the confidence level (CL), explain
what population parameter is being estimated (here, a population mean),
and should state the confidence interval (both endpoints). "We estimate with
___% confidence that the true population mean (include context of the
problem) is between and _____ (include appropriate units)."

Example:

Suppose scores on exams in statistics are normally distributed with an
unknown population mean and a population standard deviation of 3 points.
A random sample of 36 scores is taken and gives a sample mean (sample
mean score) of 68. Find a confidence interval estimate for the population
mean exam score (the mean score on all exams).

Exercise:

Problem:

Find a 90% confidence interval for the true (population) mean of
Statistics exam scores.

Solution:

e You can use technology to directly calculate the confidence
interval

e The first solution is shown step-by-step (Solution A).

e The second solution uses the TI-83, 83+ and 84+ calculators
(Solution B).

Solution A
To find the confidence interval, you need the sample mean, x, and the
EBM.

¢ ¢ = 3;n = 36; The confidence level is 90% (CL=0.90)

CL—0900s07—1 — CE 1 — 090 — 0:10

5 = 0.05 Za = 2.05

The area to the right of 2,95 is 0.05 and the area to the left of z.95 is
1-0.05=0.95

using invNorm(0.95,0,1) on the TI-83,83+,84+ calculators. This can
also be found using appropriate commands on other calculators, using
a computer, or using a probability table for the Standard Normal
distribution.

_ (ee
EBM = 1.645 ( +.) 0.8225

o— EBM 68 — 08225 — 071i
x + EBM = 68 + 0.8225 = 68.8225

The 90% confidence interval is (67.1775, 68.8225).
Solution B

Using a function of the TI-83, TI-83+ or TI-84 calculators:

Press STAT and arrow over to TESTS.

Arrow down to 7:ZInterval.

Press ENTER.

Arrow to Stats and press ENTER.

Arrow down and enter 3 for o, 68 for z , 36 for n, and .90 for C- Level.
Arrow down to Calculate and press ENTER.

The confidence interval is (to 3 decimal places) (67.178, 68.822).
Interpretation

We estimate with 90% confidence that the true population mean exam
score for all statistics students is between 67.18 and 68.82.
Explanation of 90% Confidence Level

90% of all confidence intervals constructed in this way contain the true
mean statistics exam score. For example, if we constructed 100 of these
confidence intervals, we would expect 90 of them to contain the true
population mean exam score.

Changing the Confidence Level or Sample Size

Example:Changing the Confidence Level
Exercise:

Problem:

Suppose we change the original problem by using a 95% confidence
level. Find a 95% confidence interval for the true (population) mean
Statistics exam score.

Solution:

To find the confidence interval, you need the sample mean, x, and the
EBM.

¢e ¢ = 3;n = 36; The confidence level is 95% (CL=0.95)
CL = 0.95 sow =1—-CL=1—0.95 = 0.05

x = 0.025 ze = 2,025
The area to the right of 2.925 is 0.025 and the area to the left of z.25 is
1-0.025=0.975

aed = £2,025 — 1.96

using invnorm(.975,0,1) on the TI-83,83+,84+ calculators. (This can
also be found using appropriate commands on other calculators, using
a computer, or using a probability table for the Standard Normal
distribution. )

ee ti ene ies
EBM = 1.96 (+5) 0.98

z—HBM = 68 — 0:98 = 67.02

x + EBM = 68 + 0.98 = 68.98

Interpretation

We estimate with 95 % confidence that the true population mean for all
Statistics exam scores is between 67.02 and 68.98.

Explanation of 95% Confidence Level

95% of all confidence intervals constructed in this way contain the true
value of the population mean statistics exam score.

Comparing the results

The 90% confidence interval is (67.18, 68.82). The 95% confidence
interval is (67.02, 68.98). The 95% confidence interval is wider. If you
look at the graphs, because the area 0.95 is larger than the area 0.90, it
makes sense that the 95% confidence interval is wider.

0.025 0.95 0.025
0.05 0.90 0.05

Summary: Effect of Changing the Confidence Level

e Increasing the confidence level increases the error bound, making the
confidence interval wider.

e Decreasing the confidence level decreases the error bound, making
the confidence interval narrower.

Example:Changing the Sample Size:

Suppose we change the original problem to see what happens to the error
bound if the sample size is changed.

Exercise:

Problem:

Leave everything the same except the sample size. Use the original
90% confidence level. What happens to the error bound and the
confidence interval if we increase the sample size and use n=100
instead of n=36? What happens if we decrease the sample size to
n=25 instead of n=36?

e x= 68

- EBM = 22- (=)

e o = 3; The confidence level is 90% (CL=0.90) ;
za = 295 = 1.645

Solution:

If we increase the sample size n to 100, we decrease the error bound.

i a (acon (ee Ce)
When n = 100: EBM = zg - (2) = 1.645 - (—2- ) = 0.4935

Solution:

If we decrease the sample size n to 25, we increase the error bound.

When 25 BN eee (=) = 1.645. (2) — 0.987
5 Jn /25

Summary: Effect of Changing the Sample Size

e Increasing the sample size causes the error bound to decrease, making
the confidence interval narrower.

e Decreasing the sample size causes the error bound to increase, making
the confidence interval wider.

Working Backwards to Find the Error Bound or Sample Mean

Working Bacwards to find the Error Bound or the Sample Mean
When we calculate a confidence interval, we find the sample mean and
calculate the error bound and use them to calculate the confidence interval.
But sometimes when we read statistical studies, the study may state the
confidence interval only. If we know the confidence interval, we can work
backwards to find both the error bound and the sample mean.

Finding the Error Bound

e From the upper value for the interval, subtract the sample mean
e OR, From the upper value for the interval, subtract the lower value.
Then divide the difference by 2.

Finding the Sample Mean

e Subtract the error bound from the upper value of the confidence
interval
e OR, Average the upper and lower endpoints of the confidence interval

Notice that there are two methods to perform each calculation. You can
choose the method that is easier to use with the information you know.

Example:

Suppose we know that a confidence interval is (67.18, 68.82) and we want
to find the error bound. We may know that the sample mean is 68. Or
perhaps our source only gave the confidence interval and did not tell us the
value of the the sample mean.

Calculate the Error Bound:

e If we know that the sample mean is 68: EBM = 68.82 — 68 = 0.82

e If we don't know the sample mean: EBM = eee OZ

Calculate the Sample Mean:

e If we know the error bound: x = 68.82 — 0.82 = 68

e If we don't know the error bound: z = ee) — 68

Calculating the Sample Size n

If researchers desire a specific margin of error, then they can use the error
bound formula to calculate the required sample size.

The error bound formula for a population mean when the population

standard deviation is known is EBM = zz - Vn
oes 22 :
The formula for sample size isn = aye , found by solving the error

bound formula for n

In this formula, z is Za, corresponding to the desired confidence level. A

researcher planning a study who wants a specified confidence level and
error bound can use this formula to calculate the size of the sample needed
for the study.

Example:

The population standard deviation for the age of Foothill College students
is 15 years. If we want to be 95% confident that the sample mean age is
within 2 years of the true population mean age of Foothill College students
, how many randomly selected Foothill College students must be
surveyed?

e From the problem, we know that 0 = 15 and EBM=2

° Z= Z 25 = 1.96, because the confidence level is 95%.

n= a — LCs =216.09 using the sample size equation.

e Use n = 217: Always round the answer UP to the next higher integer
to ensure that the sample size is large enough.

Therefore, 217 Foothill College students should be surveyed in order to be

95% confident that we are within 2 years of the true population mean age
of Foothill College students.

**With contributions from Roberta Bloom

Glossary
Confidence Interval (CI)
An interval estimate for an unknown population parameter. This

depends on:

e The desired confidence level.

e Information that is known about the distribution (for example,
known standard deviation).
e The sample and its size.

Confidence Level (CL)
The percent expression for the probability that the confidence interval
contains the true population parameter. For example, if the CL = 90%
, then in 90 out of 100 samples the interval estimate will enclose the
true population parameter.

Error Bound for a Population Mean (EBM)
The margin of error. Depends on the confidence level, sample size, and
known or estimated population standard deviation.

Confidence Interval, Single Population Mean, Standard Deviation
Unknown, Student's-t

Confidence Interval, Single Population Mean, Population Standard
Deviation Unknown, Student-t is part of the collection col10555 written by
Barbara Illowsky and Susan Dean with contributions from Roberta Bloom.

In practice, we rarely know the population standard deviation. In the past,
when the sample size was large, this did not present a problem to
statisticians. They used the sample standard deviation s as an estimate for 7
and proceeded as before to calculate a confidence interval with close
enough results. However, statisticians ran into problems when the sample
size was small. A small sample size caused inaccuracies in the confidence
interval.

William S. Gossett (1876-1937) of the Guinness brewery in Dublin, Ireland
ran into this problem. His experiments with hops and barley produced very
few samples. Just replacing o with s did not produce accurate results when
he tried to calculate a confidence interval. He realized that he could not use
a normal distribution for the calculation; he found that the actual
distribution depends on the sample size. This problem led him to "discover"
what is called the Student's-t distribution. The name comes from the fact
that Gosset wrote under the pen name "Student."

Up until the mid 1970s, some statisticians used the normal distribution
approximation for large sample sizes and only used the Student's-t
distribution for sample sizes of at most 30. With the common use of
graphing calculators and computers, the practice is to use the Student's-t
distribution whenever s is used as an estimate for o.

If you draw a simple random sample of size n from a population that has
approximately a normal distribution with mean pw and unknown population
x=
Ca
follow a Student's-t distribution with n — 1 degrees of freedom. The t-
score has the same interpretation as the z-score. It measures how far z is
from its mean py. For each sample size n, there is a different Student's-t
distribution.

standard deviation o and calculate the t-score t = , then the t-scores

The degrees of freedom, m — 1, come from the calculation of the sample
standard deviation s. In Chapter 2, we used n deviations (x — x values) to
calculate s. Because the sum of the deviations is 0, we can find the last
deviation once we know the other n — 1 deviations. The other n — 1
deviations can change or vary freely. We call the number n — 1 the
degrees of freedom (df).

Properties of the Student's-t Distribution

e The graph for the Student's-t distribution is similar to the Standard
Normal curve.

e The mean for the Student's-t distribution is 0 and the distribution is
symmetric about 0.

e The Student's-t distribution has more probability in its tails than the
Standard Normal distribution because the spread of the t distribution is
greater than the spread of the Standard Normal. So the graph of the
Student's-t distribution will be thicker in the tails and shorter in the
center than the graph of the Standard Normal distribution.

e The exact shape of the Student's-t distribution depends on the "degrees
of freedom". As the degrees of freedom increases, the graph Student's-t
distribution becomes more like the graph of the Standard Normal
distribution.

e The underlying population of individual observations is assumed to be
normally distributed with unknown population mean py and unknown
population standard deviation o. The size of the underlying population
is generally not relevant unless it is very small. If it is bell shaped
(normal) then the assumption is met and doesn't need discussion.
Random sampling is assumed but it is a completely separate
assumption from normality.

Calculators and computers can easily calculate any Student's-t probabilities.
The TI-83,83+,84+ have a tcdf function to find the probability for given
values of t. The grammar for the tcdf command is tcdf(lower bound, upper
bound, degrees of freedom). However for confidence intervals, we need to
use inverse probability to find the value of t when we know the probability.

For the TI-84+ you can use the invl command on the DISTRibution menu.
The invT command works similarly to the invnorm. The invT command

requires two inputs: invT(area to the left, degrees of freedom) The output
is the t-score that corresponds to the area we specified.

The TI-83 and 83+ do not have the invT command. (The TI-89 has an
inverse T command.)

A probability table for the Student's-t distribution can also be used. The
table gives t-scores that correspond to the confidence level (column) and
degrees of freedom (row). (The TI-86 does not have an invT program or
command, so if you are using that calculator, you need to use a probability
table for the Student's-t distribution.) When using t-table, note that some
tables are formatted to show the confidence level in the column headings,
while the column headings in some tables may show only corresponding
area in one or both tails.

A Student's-t table (See the Table of Contents 15. Tables) gives t-scores
given the degrees of freedom and the right-tailed probability. The table is
very limited. Calculators and computers can easily calculate any
Student's-t probabilities.

The notation for the Student's-t distribution is (using T as the random
variable) is

e T'~tg¢ where df = n — 1.

e For example, if we have a sample of size n=20 items, then we calculate
the degrees of freedom as df=n—1=20-1=19 and we write the
distribution as T' ~ ty9

If the population standard deviation is not known, the error bound for
a population mean is:

EBM = ts - (—)
¢ ta is the t-score with area to the right equal ers

e use df = n — 1 degrees of freedom
e s =sample standard deviation

The format for the confidence interval is:

(« — EBM, z + EBM).

The TI-83, 83+ and 84 calculators have a function that calculates the
confidence interval directly. To get to it,

Press STAT

Arrow over to TESTS.

Arrow down to 8: TInterval and press ENTER (or just press 8).

Example:
Exercise:

Problem:

Suppose you do a study of acupuncture to determine how effective it
is in relieving pain. You measure sensory rates for 15 subjects with
the results given below. Use the sample data to construct a 95%
confidence interval for the mean sensory rate for the population
(assumed normal) from which you took the data.

The solution is shown step-by-step and by using the TI-83, 83+ and
84+ calculators.
8.6 9.4 7.9 6.8 8.3 7.3 9.2 9.6 8.7 11.4 10.3 5.4 8.1 5.5 6.9

Solution:

e You can use technology to directly calculate the confidence
interval.

e The first solution is step-by-step (Solution A).

e The second solution uses the Ti-83+ and Ti-84 calculators
(Solution B).

Solution A
To find the confidence interval, you need the sample mean, x, and the
EBM.

ie ee PJAUT s— 10722 UG

dei = ie
CL=0.95 so a=1—CL=1—0.95 = 0.05
2=0.025 te =toos

The area to the right of ¢ 925 is 0.025 and the area to the left of ¢ 25 is
1-0.025=0.975

ta =¢ 995 = 2.14 using invT(.975,14) on the TI-84+ calculator.

EBM = 2.14: (1822) — 0,924
/15

x — EBM = 8.2267 — 0.9240 = 7.3
«4 EBM — 8.2267 -- 0.9240 — 9.15
The 95% confidence interval is (7.30, 9.15).

We estimate with 95% confidence that the true population mean
sensory rate is between 7.30 and 9.15.

Solution B
Using a function of the TI-83, TI-83+ or TI-84 calculators:

Press STAT and arrow over to TESTS.

Arrow down to 8: TInterval and press ENTER (or you can just press
8). Arrow to Data and press ENTER.

Arrow down to List and enter the list name where you put the data.
Arrow down to Freg and enter 1.

Arrow down to C- Level and enter .95

Arrow down to Calculate and press ENTER.

The 95% confidence interval is (7.3006, 9.1527)

Note: When calculating the error bound, a probability table for the
Student's-t distribution can also be used to find the value of t. The table
gives t-scores that correspond to the confidence level (column) and
degrees of freedom (row); the t-score is found where the row and column
intersect in the table.

** With contributions from Roberta Bloom

Glossary

Confidence Interval (CI)
An interval estimate for an unknown population parameter. This
depends on:

e The desired confidence level.

e Information that is known about the distribution (for example,
known standard deviation).

e The sample and its size.

Confidence Level (CL)
The percent expression for the probability that the confidence interval
contains the true population parameter. For example, if the CL = 90%
, then in 90 out of 100 samples the interval estimate will enclose the
true population parameter.

Degrees of Freedom (df)
The number of objects in a sample that are free to vary.

Error Bound for a Population Mean (EBM)
The margin of error. Depends on the confidence level, sample size, and
known or estimated population standard deviation.

Normal Distribution

A continuous random variable (RV) with pdf

f(x) = TE e~(=-1)*/20° where ps is the mean of the distribution and

o is the standard deviation. Notation: X ~ N(y, 0). If u = 0 and
ao = 1, the RV is called the standard normal distribution.

Standard Deviation
A number that is equal to the square root of the variance and measures
how far data values are from their mean. Notation: s for sample
standard deviation and o for population standard deviation.

Student's-t Distribution
Investigated and reported by William S. Gossett in 1908 and published
under the pseudonym Student. The major characteristics of the random
variable (RV) are:

e It is continuous and assumes any real values.

e The pdf is symmetrical about its mean of zero. However, it is
more spread out and flatter at the apex than the normal
distribution.

e It approaches the standard normal distribution as n gets larger.

e There is a "family" of t distributions: every representative of the
family is completely defined by the number of degrees of
freedom which is one less than the number of data.

Confidence Interval for a Population Proportion

Confidence Interval for a Population Proportion is part of the collection
col10555 written by Barbara Illowsky and Susan Dean with contributions
from Roberta Bloom.

During an election year, we see articles in the newspaper that state
confidence intervals in terms of proportions or percentages. For example, a
poll for a particular candidate running for president might show that the
candidate has 40% of the vote within 3 percentage points. Often, election
polls are calculated with 95% confidence. So, the pollsters would be 95%
confident that the true proportion of voters who favored the candidate
would be between 0.37 and 0.43 : (0.40 — 0.03, 0.40 + 0.03).

Investors in the stock market are interested in the true proportion of stocks
that go up and down each week. Businesses that sell personal computers are
interested in the proportion of households in the United States that own
personal computers. Confidence intervals can be calculated for the true
proportion of stocks that go up or down each week and for the true
proportion of households in the United States that own personal computers.

The procedure to find the confidence interval, the sample size, the error
bound, and the confidence level for a proportion is similar to that for the
population mean. The formulas are different.

How do you know you are dealing with a proportion problem? First, the
underlying distribution is binomial. (There is no mention of a mean or
average.) If X is a binomial random variable, then X~B(n, p) where n =
the number of trials and p = the probability of a success. To form a
proportion, take X, the random variable for the number of successes and
divide it by n, the number of trials (or the sample size). The random
variable P/ (read "P prime") is that proportion,

Pr=

n

(Sometimes the random variable is denoted as P, read "P hat".)

When n is large and p is not close to 0 or 1, we can use the normal
distribution to approximate the binomial.

X~ N(n-p,./n-p-q)

If we divide the random variable by n, the mean by n, and the standard
deviation by n, we get a normal distribution of proportions with P/, called
the estimated proportion, as the random variable. (Recall that a proportion =
the number of successes divided by n.)

~ = Pr~ N( 32 wrt)

n? n

Using algebra to simplify : “** = ,/2%

P? follows a normal distribution for proportions: P/ ~ N (», Ba )

The confidence interval has the form (p/—EBP, p/+EBP).

pi= +,

p! = the estimated proportion of successes (p/ is a point estimate for p,
the true proportion)

zx = the number of successes.
n = the size of the sample

The error bound for a proportion is

EBP = ze - jet where q/= 1 — pi

This formula is similar to the error bound formula for a mean, except that
the "appropriate standard deviation" is different. For a mean, when the
population standard deviation is known, the appropriate standard deviation

that we use is Vm For a proportion, the appropriate standard deviation is
Pd

Vit

However, in the error bound formula, we use 4/ Be as the standard

deviation, instead of / Pe

However, in the error bound formula, the standard deviation is ,/ Pe.

In the error bound formula, the sample proportions p/ and q/ are
estimates of the unknown population proportions p and gq. The estimated
proportions p/ and q/ are used because p and q are not known. p/ and q/ are
calculated from the data. p/ is the estimated proportion of successes. q/ is
the estimated proportion of failures.

The confidence interval can only be used if the number of successes np/
and the number of failures nq/ are both larger than 5.

Note:For the normal distribution of proportions, the z-score formula is as
follows.

If Pr~ N (», / Ea) then the z-score formula is z = 2&2

Example:
Exercise:

Problem:

Suppose that a market research firm is hired to estimate the percent of
adults living in a large city who have cell phones. 500 randomly
selected adult residents in this city are surveyed to determine whether
they have cell phones. Of the 500 people surveyed, 421 responded yes
- they own cell phones. Using a 95% confidence level, compute a
confidence interval estimate for the true proportion of adults residents
of this city who have cell phones.

Solution

e You can use technology to directly calculate the confidence
interval.

e The first solution is step-by-step (Solution A).

e The second solution uses a function of the TI-83, 83+ or 84
calculators (Solution B).

Solution:

Let X = the number of people in the sample who have cell phones. X

is binomial. X ~ B(500, =~).

To calculate the confidence interval, you must find p/, g/, and EBP.
nm = 500 zx = the number of successes = 421

WSS e SIL

pl= 0.842 is the sample proportion; this is the point estimate of the
population proportion.

gi= 1 — pl= 1 — 0.842 = 0.158

Since CL = 0.95, then
a= 1 CL = 1 — 0,95 = 0:05 3 = 0.025.

Then Za = 2.025 = 1.96

Use the TI-83, 83+ or 84+ calculator command invNorm(0.975,0,1) to
find z.925. Remember that the area to the right of 2,995 is 0.025 and the
area to the left of 29.925 is 0.975. This can also be found using
appropriate commands on other calculators, using a computer, or
using a Standard Normal probability table.

EBP = 2g -«/ ## = 1.96. 4/ S82) 00) — 0.032
2 n

pi DE — 0.642 0052 — 0.6
pI+EBP = 0.842 + 0.032 = 0.874

The confidence interval for the true binomial population proportion is
(p!—EBP, p/+EBP) =(0.810, 0.874).

Interpretation
We estimate with 95% confidence that between 81% and 87.4% of all
adult residents of this city have cell phones.

Explanation of 95% Confidence Level

95% of the confidence intervals constructed in this way would contain
the true value for the population proportion of all adult residents of
this city who have cell phones.

Solution:

Using a function of the TI-83, 83+ or 84 calculators:

Press STAT and arrow over to TESTS.

Arrow down to A:1-PropZint. Press ENTER.
Arrow down to x and enter 421.

Arrow down to n and enter 500.

Arrow down to C-Level and enter .95.

Arrow down to Calculate and press ENTER.
The confidence interval is (0.81003, 0.87397).

Example:
Exercise:

Problem:

For a class project, a political science student at a large university
wants to estimate the percent of students that are registered voters. He
surveys 500 students and finds that 300 are registered voters.
Compute a 90% confidence interval for the true percent of students
that are registered voters and interpret the confidence interval.

Solution:

e You can use technology to directly calculate the confidence
interval.

e The first solution is step-by-step (Solution A).

e The second solution uses a function of the TI-83, 83+ or 84
calculators (Solution B).

Solution A
xe = 300 andar— 500.

eS = Savi

gi= 1 — pf= 1 — 0.600 = 0.400

Since CL = 0.90, then
a=1—CL=1-—0.90 = 0.10 5 = 0.05.

Ze = 2.05 = 1.645

Use the TI-83, 83+ or 84+ calculator command invNorm(0.95,0,1) to
find z.95. Remember that the area to the right of z.95 is 0.05 and the
area to the left of z.95 is 0.95. This can also be found using

appropriate commands on other calculators, using a computer, or
using a Standard Normal probability table.

EBP = ze - / 22 = 1.645 - 4/ 20) _ 9.036

p!—EBP = 0.60 — 0.036 = 0.564

pI+EBP = 0.60 + 0.036 = 0.636

The confidence interval for the true binomial population proportion is
(p!—EBP, p/+EBP) =(0.564, 0.636).
Interpretation:

e We estimate with 90% confidence that the true percent of all
students that are registered voters is between 56.4% and 63.6%.

e Alternate Wording: We estimate with 90% confidence that
between 56.4% and 63.6% of ALL students are registered voters.

Explanation of 90% Confidence Level

90% of all confidence intervals constructed in this way contain the
true value for the population percent of students that are registered
voters.

Solution B
Using a function of the TI-83, 83+ or 84 calculators:

Press STAT and arrow over to TESTS.

Arrow down to A:1-PropZint. Press ENTER.
Arrow down to x and enter 300.

Arrow down to n and enter 500.

Arrow down to C-Level and enter .90.

Arrow down to Calculate and press ENTER.
The confidence interval is (0.564, 0.636).

Calculating the Sample Size n

If researchers desire a specific margin of error, then they can use the error
bound formula to calculate the required sample size.

The error bound formula for a population proportion is

* EBP = ze - auicle

n
e Solving for n gives you an equation for the sample size.

Example:

Suppose a mobile phone company wants to determine the current
percentage of customers aged 50+ that use text messaging on their cell
phone. How many customers aged 50+ should the company survey in order
to be 90% confident that the estimated (sample) proportion is within 3
percentage points of the true population proportion of customers aged 50+
that use text messaging on their cell phone.

Solution
From the problem, we know that EBP=0.03 (3%=0.03) and

Za = 295 = 1.645 because the confidence level is 90%

However, in order to find n , we need to know the estimated (sample)
proportion p'. Remember that q'=1-p'. But, we do not know p' yet. Since we
multiply p' and q' together, we make them both equal to 0.5 because p'q'=
(.5)(.5)=.25 results in the largest possible product. (Try other products: (.6)
(.4)=.24; (.3)(.7)=.21; (.2)(.8)=.16 and so on). The largest possible product
gives us the largest n. This gives us a large enough sample so that we can
be 90% confident that we are within 3 percentage points of the true
population proportion. To calculate the sample size n, use the formula and
make the substitutions.

ee 1.6452(.5)(.5) _
Z Pg oe = foley

Round the answer to the next higher value. The sample size should be 752
cell phone customers aged 50+ in order to be 90% confident that the

o— gives n =

estimated (sample) proportion is within 3 percentage points of the true
population proportion of all customers aged 50+ that use text messaging on
their cell phone.

**With contributions from Roberta Bloom.

Glossary

Binomial Distribution
A discrete random variable (RV) which arises from Bernoulli trials.
There are a fixed number, n, of independent trials. “Independent”
means that the result of any trial (for example, trial 1) does not affect
the results of the following trials, and all trials are conducted under the
same conditions. Under these circumstances the binomial RV X is
defined as the number of successes in 7 trials. The notation is: X~
B(n, p). The mean is js = np and the standard deviation is o = ,/npq
. The probability of exactly z successes in 7 trials is

P(X =2) = (1 )p"q"*.

Confidence Interval (CI)
An interval estimate for an unknown population parameter. This
depends on:

e The desired confidence level.

e Information that is known about the distribution (for example,
known standard deviation).

e The sample and its size.

Confidence Level (CL)
The percent expression for the probability that the confidence interval
contains the true population parameter. For example, if the CL = 90%
, then in 90 out of 100 samples the interval estimate will enclose the
true population parameter.

Error Bound for a Population Proportion(EBP)
The margin of error. Depends on the confidence level, sample size, and
the estimated (from the sample) proportion of successes.

Normal Distribution

A continuous random variable (RV) with pdf
2
f(x) = amore 20 where p is the mean of the distribution and

o is the standard deviation. Notation: X ~ N(w,o). If u = 0 and
ao = 1, the RV is called the standard normal distribution.

Summary of Formulas

Formula General form of a confidence interval

(lower value, upper value) = (point estimate — error bound, point estimate + error bound)
FormulaTo find the error bound when you know the confidence interval

error bound = upper value — point estimate OR

écror bound — upper eine fore value

FormulaSingle Population Mean, Known Standard Deviation, Normal Distribution
Use the Normal Distribution for Means EBM =72 Vn
The confidence interval has the format ( - EBM, # + EBM).

FormulaSingle Population Mean, Unknown Standard Deviation, Student's-t Distribution

Use the Student's-t Distribution with degrees of freedom df = n — 1. EBM = fz - A

FormulaSingle Population Proportion, Normal Distribution

Use the Normal Distribution for a single population proportion pl= =

EBP =z2-,/7" — pr+qr=1

The confidence interval has the format (p/—EBP, p/+-EBP).
FormulaPoint Estimates

Z is a point estimate for yz

p!/is a point estimate for p

s is a point estimate for 7

Practice 1: Confidence Intervals for Means, Known Population Standard
Deviation

Student Learning Outcomes

e The student will calculate confidence intervals for means when the
population standard deviation is known.

Given

The mean age for all Foothill College students for a recent Fall term was
33.2. The population standard deviation has been pretty consistent at 15.
Suppose that twenty-five Winter students were randomly selected. The
mean age for the sample was 30.4. We are interested in the true mean age
for Winter Foothill College students.

(http://research.fhda.edu/factbook/FH Demo _Trends/FoothillDemographic
Trends.htm

Let X = the age of a Winter Foothill College student

Calculating the Confidence Interval

Exercise:

Problem: x =

Solution:

30.4

Exercise:

Problem: n=

Solution:

25

Exercise:

Problem: 15=(insert symbol here)

Solution:

oO

Exercise:

Problem: Define the Random Variable, X , in words.
X=
Solution:

the mean age of 25 randomly selected Winter Foothill students

Exercise:

Problem: What is x estimating?
Solution:

LL
Exercise:

Problem: Is 7 ,, known?
Solution:

yes

Exercise:

Problem:

As aresult of your answer to (4), state the exact distribution to use
when calculating the Confidence Interval.

Solution:

Normal

Explaining the Confidence Interval

Construct a 95% Confidence Interval for the true mean age of Winter
Foothill College students.
Exercise:

Problem: How much area is in both tails (combined)? a = __

Solution:

0.05

Exercise:

Problem: How much area is in each tail? > =

Solution:

0.025

Exercise:
Problem: Identify the following specifications:
e a lower limit =

¢ b upper limit =
e cerror bound =

Solution:

e a24.52
° b36.28
e c5.88

Exercise:

Problem: The 95% Confidence Interval is:
Solution:

(24 be 36.28)
Exercise:
Problem:

Fill in the blanks on the graph with the areas, upper and lower
limits of the Confidence Interval, and the sample mean.

ae CL.= =
z2 | nn 2

Exercise:
Problem: In one complete sentence, explain what the interval means.

Discussion Questions

Exercise:

Problem:

Using the same mean, standard deviation and level of confidence,
suppose that n were 69 instead of 25. Would the error bound become
larger or smaller? How do you know?

Exercise:
Problem:
Using the same mean, standard deviation and sample size, how would

the error bound change if the confidence level were reduced to 90%?
Why?

Practice 2: Confidence Intervals for Means, Unknown Population Standard
Deviation

Student Learning Outcomes

e The student will calculate confidence intervals for means when the
population standard deviation is unknown.

Given

The following real data are the result of a random survey of 39 national
flags (with replacement between picks) from various countries. We are
interested in finding a confidence interval for the true mean number of
colors on a national flag. Let X = the number of colors on a national flag.

xX Freq.
1 1

2 7

3 18

4 7

5 6

Calculating the Confidence Interval

Exercise:

Problem: Calculate the following:

e at=
e bs, =
e* cn =

Solution:

e a3.26
e b1.02
e ¢39

Exercise:

Problem:

Define the Random Variable, X, in words. X =

Solution:

the mean number of colors of 39 flags

Exercise:

Problem: What is Z estimating?
Solution:

LL
Exercise:

Problem: Is 7 ,, known?

Solution:

No
Exercise:

Problem:

As aresult of your answer to (4), state the exact distribution to use
when calculating the Confidence Interval.

Solution:

t38

Confidence Interval for the True Mean Number

Construct a 95% Confidence Interval for the true mean number of colors on
national flags.
Exercise:

Problem: How much area is in both tails (combined)? a =

Solution:
0.05

Exercise:

Problem: How much area is in each tail? > =

Solution:
0.025
Exercise:
Problem: Calculate the following:

e alower limit =

¢ bupper limit =
e cerror bound =

Solution:

e a2.93
e b3.59
e 0.33

Exercise:

Problem: The 95% Confidence Interval is:
Solution:

2.93: 3.59
Exercise:

Problem:

Fill in the blanks on the graph with the areas, upper and lower limits of
the Confidence Interval and the sample mean.

= = é3 =
2 2

Exercise:
Problem: In one complete sentence, explain what the interval means.

Discussion Questions

Exercise:
Problem:
Using the same Z, s,, and level of confidence, suppose that nm were 69

instead of 39. Would the error bound become larger or smaller? How
do you know?

Exercise:
Problem:

Using the same 7, s,, and n = 39, how would the error bound change
if the confidence level were reduced to 90%? Why?

Practice 3: Confidence Intervals for Proportions

Student Learning Outcomes

e The student will calculate confidence intervals for proportions.

Given

The Ice Chalet offers dozens of different beginning ice-skating classes. All
of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 -
12, beginning ice-skating class was picked. In that class were 64 girls and
16 boys. Suppose that we are interested in the true proportion of girls, ages
8 - 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the
children in the selected class is a random sample of the population.

Estimated Distribution
Exercise:
Problem: What is being counted?
Exercise:
Problem: In words, define the Random Variable X.X =

Solution:
The number of girls, age 8-12, in the beginning ice skating class
Exercise:
Problem: Calculate the following:
eaxrt=—

e bn=
e Cc piI=

Solution:

e a64
e b80
e c0.8

Exercise:

Problem: State the estimated distribution of X.X ~

Solution:
B(80, 0.80)
Exercise:
Problem: Define a new Random Variable P/. What is p/ estimating?
Solution:

Pp
Exercise:

Problem: In words, define the Random Variable P/ . P/=
Solution:

The proportion of girls, age 8-12, in the beginning ice skating class.

Exercise:

Problem: State the estimated distribution of P/. P!~

Explaining the Confidence Interval

Construct a 92% Confidence Interval for the true proportion of girls in the
age 8 - 12 beginning ice-skating classes at the Ice Chalet.
Exercise:

Problem: How much area is in both tails (combined)? a =

Solution:
1 - 0.92 = 0.08
Exercise:

Problem: How much area is in each tail? = =

Solution:
0.04
Exercise:

Problem: Calculate the following:
e alower limit =
¢ bupper limit =
e cerror bound =

Solution:

e a0.72
e b0.88
e c0.08

Exercise:

Problem: The 92% Confidence Interval is:

Solution:

(0.72; 0.88)
Exercise:

Problem:
Fill in the blanks on the graph with the areas, upper and lower

limits of the Confidence Interval, and the sample proportion.
a

a L= a
27-—— h__ y=

Exercise:

Problem: In one complete sentence, explain what the interval means.

Discussion Questions

Exercise:
Problem:
Using the same p/ and level of confidence, suppose that n were

increased to 100. Would the error bound become larger or smaller?
How do you know?

Exercise:
Problem:
Using the same p/ and n = 80, how would the error bound change if
the confidence level were increased to 98%? Why?

Exercise:

Problem:

If you decreased the allowable error bound, why would the minimum
sample size increase (keeping the same level of confidence)?

Homework

Note:If you are using a student's-t distribution for a homework problem
below, you may assume that the underlying population is normally
distributed. (In general, you must first prove that assumption, though.)

Exercise:

Problem:

Among various ethnic groups, the standard deviation of heights is
known to be approximately 3 inches. We wish to construct a 95%
confidence interval for the mean height of male Swedes. 48 male
Swedes are surveyed. The sample mean is 71 inches. The sample
standard deviation is 2.8 inches.

eda

if. - .
ic =_—
ili s, =
ivn =

vn —1=

Oo: O,O-«O: -0

e bDefine the Random Variables X and X, in words.

¢ cWhich distribution should you use for this problem? Explain
your choice.

e dConstruct a 95% confidence interval for the population mean
height of male Swedes.

o iState the confidence interval.
© 1iSketch the graph.
o jiiCalculate the error bound.

e¢ eWhat will happen to the level of confidence obtained if 1000

male Swedes are surveyed instead of 48? Why?

Solution:

a

i71
13
112.8
iv48
v47

oO. 0: 20: O- -O

cN (71,—=)

? ./48
d

o iCI: (70.15,71.85)
o fiiEB = 0.85

Exercise:

Problem:

In six packages of “The Flintstones® Real Fruit Snacks” there were 5
Bam-Bam snack pieces. The total number of snack pieces in the six
bags was 68. We wish to calculate a 96% confidence interval for the
population proportion of Bam-Bam snack pieces.

aDefine the Random Variables X and P’, in words.

bWhich distribution should you use for this problem? Explain
your choice

cCalculate p’.

dConstruct a 96% confidence interval for the population
proportion of Bam-Bam snack pieces per bag.

o 4 State the confidence interval.
© 1iSketch the graph.

o jiiCalculate the error bound.

e eDo you think that six packages of fruit snacks yield enough data
to give accurate results? Why or why not?

Exercise:

Problem:

A random survey of enrollment at 35 community colleges across the
United States yielded the following figures (source: Microsoft
Bookshelf): 6414; 1550; 2109; 9350; 21828; 4300; 5944; 5722; 2825;
2044; 5481; 5200; 5853; 2750; 10012; 6357; 27000; 9414; 7681;
3200; 17500; 9200; 7380; 18314; 6557; 13713; 17768; 7493; 2771;
2861; 1263; 7285; 28165; 5080; 11622. Assume the underlying
population is normal.

ea

=

1
i |, —
ivn —1=

oO 0 Oo 0

e bDefine the Random Variables X and X, in words.

e¢ cWhich distribution should you use for this problem? Explain
your choice.

e dConstruct a 95% confidence interval for the population mean
enrollment at community colleges in the United States.

o iState the confidence interval.
o 1iSketch the graph.
o jiiCalculate the error bound.

e eWhat will happen to the error bound and confidence interval if
500 community colleges were surveyed? Why?

Solution:

ea
18629
116944
1135
iv34

o Oo 0 0

e Cc t34
ed

o iCI: (6244, 11,014)
o iiiEB = 2385

e elt will become smaller

Exercise:

Problem:

From a stack of IEEE Spectrum magazines, announcements for 84
upcoming engineering conferences were randomly picked. The mean
length of the conferences was 3.94 days, with a standard deviation of
1.28 days. Assume the underlying population is normal.

e aDefine the Random Variables X and X, in words.

e¢ bWhich distribution should you use for this problem? Explain
your choice.

e cConstruct a 95% confidence interval for the population mean
length of engineering conferences.

o iState the confidence interval.

o 11Sketch the graph.
© jiiCalculate the error bound.

Exercise:

Problem:

Suppose that a committee is studying whether or not there is waste of
time in our judicial system. It is interested in the mean amount of time
individuals waste at the courthouse waiting to be called for service.
The committee randomly surveyed 81 people. The sample mean was 8
hours with a sample standard deviation of 4 hours.

ea

iS.
ls, =
HS 2

ivn —1=

Oo oOo Oo Oo

e bDefine the Random Variables X and X, in words.

e¢ cWhich distribution should you use for this problem? Explain
your choice.

e dConstruct a 95% confidence interval for the population mean
time wasted.

°o aState the confidence interval.
© bSketch the graph.
o cCalculate the error bound.

e eExplain in a complete sentence what the confidence interval
means.

Solution:

ea
i8
ii4
11181
iv80

o Oo 0 0

e C tao

ed
© ICI: (7.12, 8.88)
°o 1iiEB = 0.88
Exercise:
Problem:

Suppose that an accounting firm does a study to determine the time
needed to complete one person’s tax forms. It randomly surveys 100
people. The sample mean is 23.6 hours. There is a known standard
deviation of 7.0 hours. The population distribution is assumed to be
normal.

ea

in.
lio =
liis, =
ivn = _

wn —-1l=

oOo 0 0 0

e bDefine the Random Variables X and X, in words.

e cWhich distribution should you use for this problem? Explain
your choice.

e dConstruct a 90% confidence interval for the population mean
time to complete the tax forms.

o 4State the confidence interval.
o 1iSketch the graph.
o jiiCalculate the error bound.

e elf the firm wished to increase its level of confidence and keep
the error bound the same by taking another survey, what changes
should it make?

e flIf the firm did another survey, kept the error bound the same, and
only surveyed 49 people, what would happen to the level of
confidence? Why?

¢ gSuppose that the firm decided that it needed to be at least 96%
confident of the population mean length of time to within 1 hour.
How would the number of people the firm surveys change? Why?

Exercise:

Problem:

A sample of 16 small bags of the same brand of candies was selected.
Assume that the population distribution of bag weights is normal. The
weight of each bag was then recorded. The mean weight was 2 ounces
with a standard deviation of 0.12 ounces. The population standard
deviation is known to be 0.1 ounce.

eda

C=. — = —.
ioe =. +;
1s. =
ivn =

vn —1=

OH 0 OO” OF 0

e bDefine the Random Variable X, in words.

° cDefine the Random Variable X, in words.

e dWhich distribution should you use for this problem? Explain
your choice.

e eConstruct a 90% confidence interval for the population mean
weight of the candies.

o iState the confidence interval.
© 1iSketch the graph.
o jiiCalculate the error bound.

e fConstruct a 98% confidence interval for the population mean
weight of the candies.

o 4State the confidence interval.
© 1iSketch the graph.
o jiiCalculate the error bound.

e gIn complete sentences, explain why the confidence interval in (f)
is larger than the confidence interval in (e).

¢ hin complete sentences, give an interpretation of what the interval
in (f) means.

Solution:
ea

i2

110.1
ili 0.12
iv16
vl5

oOo 0 0 0

e bthe weight of 1 small bag of candies
e cthe mean weight of 16 small bags of candies
¢ dN (2, iG)

*e

o i CTI: (1.96, 2.04)
o iii EB = 0.04

o 1 CI: (1.94, 2.06)
° iii EB = 0.06

Exercise:

Problem:

A pharmaceutical company makes tranquilizers. It is assumed that the
distribution for the length of time they last is approximately normal.
Researchers in a hospital used the drug on a random sample of 9
patients. The effective period of the tranquilizer for each patient (in
hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.

eda

a
lis, =
1 —

ivn —1=

oO oOo © Oo

e bDefine the Random Variable X, in words.

¢ cDefine the Random Variable X, in words.

e dWhich distribution should you use for this problem? Explain
your choice.

e eConstruct a 95% confidence interval for the population mean
length of time.

o 4State the confidence interval.
o 1iSketch the graph.
o jiiCalculate the error bound.

e fWhat does it mean to be “95% confident” in this problem?
Exercise:
Problem:

Suppose that 14 children were surveyed to determine how long they
had to use training wheels. It was revealed that they used them an
average of 6 months with a sample standard deviation of 3 months.
Assume that the underlying population distribution is normal.

ea

It = = =

1.
V1)
ivn —1=

o Oo 0 0

e bDefine the Random Variable X, in words.

° cDefine the Random Variable X, in words.

e dWhich distribution should you use for this problem? Explain
your choice.

e eConstruct a 99% confidence interval for the population mean
length of time using training wheels.

o iState the confidence interval.
© 1iSketch the graph.
o j1iCalculate the error bound.

e fWhy would the error bound change if the confidence level was
lowered to 90%?

Solution:

ea
i6
113
11114
iv13

Oo oO 0 -O

e bthe time for a child to remove his training wheels

e cthe mean time for 14 children to remove their training wheels.
e dfs

ee

© iCI: (3.58, 8.42)
o iiiEB = 2.42

Exercise:

Problem:

Insurance companies are interested in knowing the population percent
of drivers who always buckle up before riding in a car.

e aWhen designing a study to determine this population proportion,
what is the minimum number you would need to survey to be
95% confident that the population proportion is estimated to
within 0.03?

e blIf it was later determined that it was important to be more than
95% confident and a new survey was commissioned, how would
that affect the minimum number you would need to survey?
Why?

Exercise:

Problem:

Suppose that the insurance companies did do a survey. They randomly
surveyed 400 drivers and found that 320 claimed to always buckle up.
We are interested in the population proportion of drivers who claim to
always buckle up.

ea

10) 1x2 —
oa |: <n

° jiip’ =

e bDefine the Random Variables X and P’, in words.

¢ cWhich distribution should you use for this problem? Explain
your choice.

e dConstruct a 95% confidence interval for the population
proportion that claim to always buckle up.

o 4State the confidence interval.
© iSketch the graph.
o jiiCalculate the error bound.

e elf this survey were done by telephone, list 3 difficulties the
companies might have in obtaining random results.

Solution:
ea

© 1320
° ii 400
° 1110.80

(0.80) (0.20)
bd cv (0.80, y 25 |

ed

© iCI: (0.76, 0.84)
o iii EB = 0.04

Exercise:

Problem:

Unoccupied seats on flights cause airlines to lose revenue. Suppose a
large airline wants to estimate its mean number of unoccupied seats
per flight over the past year. To accomplish this, the records of 225
flights are randomly selected and the number of unoccupied seats is
noted for each of the sampled flights. The sample mean is 11.6 seats
and the sample standard deviation is 4.1 seats.

ea

— — ae
lis, =
Wie SW __._.

ivn —1=

o Oo 0 0

e bDefine the Random Variables X and X, in words.

¢ cWhich distribution should you use for this problem? Explain
your choice.

e dConstruct a 92% confidence interval for the population mean
number of unoccupied seats per flight.

o 4State the confidence interval.
© 1iSketch the graph.
© tii Calculate the error bound.

Exercise:

Problem:

According to a recent survey of 1200 people, 61% feel that the
president is doing an acceptable job. We are interested in the
population proportion of people who feel the president is doing an
acceptable job.

e aDefine the Random Variables X and P’, in words.

¢ bWhich distribution should you use for this problem? Explain
your choice.

e cConstruct a 90% confidence interval for the population
proportion of people who feel the president is doing an acceptable
job.

o 4State the confidence interval.
o 1iSketch the graph.
o jiiCalculate the error bound.

Solution:

(0.61)(0.39)

ec

© iCI: (0.59, 0.63)
o iii EB = 0.02

Exercise:

Problem:

A survey of the mean amount of cents off that coupons give was done
by randomly surveying one coupon per page from the coupon sections
of a recent San Jose Mercury News. The following data were
collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50;
A40¢; 65¢; 40¢. Assume the underlying distribution is approximately
normal.

eda

| i —
1.33 =
liin = _
ivn —1=

Oo Oo 0 0

e bDefine the Random Variables X and X, in words.

¢ cWhich distribution should you use for this problem? Explain
your choice.

e dConstruct a 95% confidence interval for the population mean
worth of coupons.

o 4State the confidence interval.
o 1iSketch the graph.
o jiiCalculate the error bound.

e elf many random samples were taken of size 14, what percent of
the confident intervals constructed should contain the population
mean worth of coupons? Explain why.

Exercise:

Problem:

An article regarding interracial dating and marriage recently appeared
in the Washington Post. Of the 1709 randomly selected adults, 315
identified themselves as Latinos, 323 identified themselves as blacks,
254 identified themselves as Asians, and 779 identified themselves as
whites. In this survey, 86% of blacks said that their families would
welcome a white person into their families. Among Asians, 77%
would welcome a white person into their families, 71% would
welcome a Latino, and 66% would welcome a black person.

e aWe are interested in finding the 95% confidence interval for the
percent of all black families that would welcome a white person
into their families. Define the Random Variables X and P’, in
words.

e bWhich distribution should you use for this problem? Explain
your choice.

e cConstruct a 95% confidence interval

o iState the confidence interval.

© 1iSketch the graph.
o jiiCalculate the error bound.

Solution:

(0.86)(0.14)

ec

o iCI: (0.823, 0.898)
© fii EB = 0.038

Exercise:

Problem:Refer to the problem above.

e aConstruct three 95% confidence intervals.

o iPercent of all Asians that would welcome a white person
into their families.

o iiPercent of all Asians that would welcome a Latino into
their families.

© iiiPercent of all Asians that would welcome a black person
into their families.

e bEven though the three point estimates are different, do any of the
confidence intervals overlap? Which?

e cFor any intervals that do overlap, in words, what does this imply
about the significance of the differences in the true proportions?

e dFor any intervals that do not overlap, in words, what does this
imply about the significance of the differences in the true
proportions?

Exercise:
Problem:
A camp director is interested in the mean number of letters each child
sends during his/her camp session. The population standard deviation

is known to be 2.5. A survey of 20 campers is taken. The mean from
the sample is 7.9 with a sample standard deviation of 2.8.

ea

oO O° Of OO
oe ple e e
pe
vA)
8
|

¢ bDefine the Random Variables X and X, in words.
e¢ cWhich distribution should you use for this problem? Explain
your choice.

e dConstruct a 90% confidence interval for the population mean
number of letters campers send home.

o 4State the confidence interval.
© 1iSketch the graph.
o jiiCalculate the error bound.

e eWhat will happen to the error bound and confidence interval if
500 campers are surveyed? Why?

Solution:
ea

17.9
2.5
iii 2.8
iv 20
v 19

oOo 0 0 0

‘ 2.5
c N(7.9, 7)
ed

o i CI: (6.98, 8.82)
© li EB: 0.92
Exercise:
Problem:
Stanford University conducted a study of whether running is healthy
for men and women over age 50. During the first eight years of the
study, 1.5% of the 451 members of the 50-Plus Fitness Association

died. We are interested in the proportion of people over 50 who ran
and died in the same eight-year period.

e aDefine the Random Variables X and P’, in words.

¢ bWhich distribution should you use for this problem? Explain
your choice.

e cConstruct a 97% confidence interval for the population
proportion of people over 50 who ran and died in the same eight—
year period.

o iState the confidence interval.
o 1iSketch the graph.
o jiiCalculate the error bound.

e d Explain what a “97% confidence interval” means for this study.

Exercise:

Problem:

In a recent sample of 84 used cars sales costs, the sample mean was
$6425 with a standard deviation of $3156. Assume the underlying
distribution is approximately normal.

e aWhich distribution should you use for this problem? Explain
your choice.

e bDefine the Random Variable X, in words.

e cConstruct a 95% confidence interval for the population mean
cost of a used car.

o iState the confidence interval.
© 11Sketch the graph.
o j1iCalculate the error bound.

e dExplain what a “95% confidence interval” means for this study.

Solution:

° atg3
e bmean cost of 84 used cars
re

© iCI: (5740.10, 7109.90)
© iii EB = 684.90

Exercise:

Problem:

A telephone poll of 1000 adult Americans was reported in an issue of
Time Magazine. One of the questions asked was “What is the main
problem facing the country?” 20% answered “crime”. We are
interested in the population proportion of adult Americans who feel
that crime is the main problem.

aDefine the Random Variables X and P’, in words.

bWhich distribution should you use for this problem? Explain
your choice.

cConstruct a 95% confidence interval for the population
proportion of adult Americans who feel that crime is the main
problem.

o iState the confidence interval.
© 1iSketch the graph.
o jiiCalculate the error bound.

dSuppose we want to lower the sampling error. What is one way
to accomplish that?

eThe sampling error given by Yankelovich Partners, Inc. (which
conducted the poll) is + 3%. In 1-3 complete sentences, explain
what the + 3% represents.

Exercise:

Problem:

Refer to the above problem. Another question in the poll was “[How
much are] you worried about the quality of education in our schools?”
63% responded “a lot”. We are interested in the population proportion
of adult Americans who are worried a lot about the quality of
education in our schools.

1. Define the Random Variables X and P’, in words.

2. Which distribution should you use for this problem? Explain your
choice.

3. Construct a 95% confidence interval for the population proportion
of adult Americans worried a lot about the quality of education in
our schools.

o iState the confidence interval.
© 1iSketch the graph.
o jiiCalculate the error bound.

4. The sampling error given by Yankelovich Partners, Inc. (which
conducted the poll) is + 3%. In 1-3 complete sentences, explain
what the + 3% represents.

Solution:
(0.63)(0.37)
ec

© iCI: (0.60, 0.66)
o iii EB = 0.03

Exercise:

Problem:

Six different national brands of chocolate chip cookies were randomly
selected at the supermarket. The grams of fat per serving are as
follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is
approximately normal.

e aCalculate a 90% confidence interval for the population mean
grams of fat per serving of chocolate chip cookies sold in
supermarkets.

o iState the confidence interval.
© 1iSketch the graph.
o jiiCalculate the error bound.

e bIf you wanted a smaller error bound while keeping the same
level of confidence, what should have been changed in the study
before it was done?

e cGo to the store and record the grams of fat per serving of six
brands of chocolate chip cookies.

e dCalculate the mean.

e els the mean within the interval you calculated in part (a)? Did
you expect it to be? Why or why not?

Exercise:
Problem:
A confidence interval for a proportion is given to be (— 0.22, 0.34).

Why doesn’t the lower limit of the confidence interval make practical
sense? How should it be changed? Why?

Try these multiple choice questions.
The next three problems refer to the following: According to a Field

Poll, 79% of California adults (actual results are 400 out of 506 surveyed)
feel that “education and our schools” is one of the top issues facing

California. We wish to construct a 90% confidence interval for the true
proportion of California adults who feel that education and the schools is
one of the top issues facing California. (Source:
http://field.com/fieldpollonline/subscribers/)

Exercise:

Problem:A point estimate for the true population proportion is:

e A0.90
e B1.27
e €0.79
e D400

Solution:

C

Exercise:

Problem:A 90% confidence interval for the population proportion is:

¢ A(0.761, 0.820)
¢ B(0.125, 0.188)
° C(0.755, 0.826)
¢ D(0.130, 0.183)

Solution:

A

Exercise:

Problem:The error bound is approximately

e Al.581
e BO.791

e €0.059
e DO0.030

Solution:

D

The next two problems refer to the following:

A quality control specialist for a restaurant chain takes a random sample of
size 12 to check the amount of soda served in the 16 oz. serving size. The
sample mean is 13.30 with a sample standard deviation of 1.55. Assume the
underlying population is normally distributed.

Exercise:

Problem:

Find the 95% Confidence Interval for the true population mean for the
amount of soda served.

e A(12.42, 14.18)
e B(12.32, 14.29)
e¢ €(12.50, 14.10)
e DImpossible to determine

Solution:

B

Exercise:

Problem: What is the error bound?

e A0.87
e B1.98
e €0.99
e D1.74

Solution:

C
Exercise:

Problem:

What is meant by the term “90% confident” when constructing a
confidence interval for a mean?

e Alf we took repeated samples, approximately 90% of the samples
would produce the same confidence interval.

e BIf we took repeated samples, approximately 90% of the
confidence intervals calculated from those samples would contain
the sample mean.

¢ ClIf we took repeated samples, approximately 90% of the
confidence intervals calculated from those samples would contain
the true value of the population mean.

e DIf we took repeated samples, the sample mean would equal the
population mean in approximately 90% of the samples.

Solution:

C

The next two problems refer to the following:

Five hundred and eleven (511) homes in a certain southern California
community are randomly surveyed to determine if they meet minimal
earthquake preparedness recommendations. One hundred seventy-three
(173) of the homes surveyed met the minimum recommendations for
earthquake preparedness and 338 did not.

Exercise:

Problem:

Find the Confidence Interval at the 90% Confidence Level for the true
population proportion of southern California community homes
meeting at least the minimum recommendations for earthquake
preparedness.

¢ A(0.2975, 0.3796)
¢ B(0.6270, 6959)

¢ C(0.3041, 0.3730)
¢ D(0.6204, 0.7025)

Solution:

C
Exercise:

Problem:

The point estimate for the population proportion of homes that do not
meet the minimum recommendations for earthquake preparedness is:

e A0.6614
¢ B0.3386
e C173
e D338

Solution:

A

Review

The next three problems refer to the following situation: Suppose that a
sample of 15 randomly chosen people were put on a special weight loss
diet. The amount of weight lost, in pounds, follows an unknown distribution
with mean equal to 12 pounds and standard deviation equal to 3 pounds.
Assume that the distribution for the weight loss is normal.

Exercise:

Problem:

To find the probability that the mean amount of weight lost by 15
people is no more than 14 pounds, the random variable should be:

A The number of people who lost weight on the special weight
loss diet

BThe number of people who were on the diet

CThe mean amount of weight lost by 15 people on the special
weight loss diet

DThe total amount of weight lost by 15 people on the special
weight loss diet

Solution:

C

Exercise:

Problem:Find the probability asked for in the previous problem.

Solution:

0.9951

Exercise:

Problem:

Find the 90th percentile for the mean amount of weight lost by 15
people.

Solution:

12,99

The next three questions refer to the following situation: The time of
occurrence of the first accident during rush-hour traffic at a major
intersection is uniformly distributed between the three hour interval 4 p.m.
to 7 p.m. Let X = the amount of time (hours) it takes for the first accident to
occur.

e So, if an accident occurs at 4 p.m., the amount of time, in hours, it took
for the accident to occur is

e pL —

e o —

Exercise:

Problem:

What is the probability that the time of occurrence is within the first
half-hour or the last hour of the period from 4 to 7 p.m.?

e ACannot be determined from the information given

B
e C
D

wl Rr] RoO|e

Solution:

G

Exercise:

Problem:The 20th percentile occurs after how many hours?

e A0.20
e B 0.60
e C 0.50
° Dil

Solution:

B
Exercise:
Problem:
Assume Ramon has kept track of the times for the first accidents to

occur for 40 different days. Let C’' = the total cumulative time. Then C’
follows which distribution?

e AU(0,3)

°B Exp(+)

° C N(60,5.477)

* D N(1.5,0.01875)

Solution:

C
Exercise:

Problem:

Using the information in question #6, find the probability that the total
time for all first accidents to occur is more than 43 hours.

Solution:

0.9990

The next two questions refer to the following situation: The length of
time a parent must wait for his children to clean their rooms is uniformly
distributed in the time interval from 1 to 15 days.

Exercise:

Problem:

How long must a parent expect to wait for his children to clean their
rooms?

e A 8 days
e B3 days
e C 14 days
e D6 days

Solution:

A
Exercise:

Problem:

What is the probability that a parent will wait more than 6 days given
that the parent has already waited more than 3 days?

¢ A0.5174
e B0.0174
e €C 0.7500
¢ D0.2143

Solution:

‘s

The next five problems refer to the following study: Twenty percent of
the students at a local community college live in within five miles of the
campus. Thirty percent of the students at the same community college
receive some kind of financial aid. Of those who live within five miles of
the campus, 75% receive some kind of financial aid.

Exercise:

Problem:

Find the probability that a randomly chosen student at the local
community college does not live within five miles of the campus.

¢ A 80%
¢ B20%
¢ C 30%
e D Cannot be determined

Solution:

A
Exercise:
Problem:
Find the probability that a randomly chosen student at the local

community college lives within five miles of the campus or receives
some kind of financial aid.

¢ A50%
¢ B35%
¢ C 27.5%
¢ D75%

Solution:

B

Exercise:

Problem:

Based upon the above information, are living in student housing within
five miles of the campus and receiving some kind of financial aid
mutually exclusive?

e A Yes
e BNo
e C Cannot be determined

Solution:

B
Exercise:

Problem:
The interest rate charged on the financial aid is data.

e A quantitative discrete

e B quantitative continuous
¢ Cqualitative discrete

¢ D qualitative

Solution:

B
Exercise:

Problem:

What follows is information about the students who receive financial
aid at the local community college.

e ist quartile = $250

e 2nd quartile = $700
e 3rd quartile = $1200

(These amounts are for the school year.) If a sample of 200 students is
taken, how many are expected to receive $250 or more?

e A50

e B250

e C 150

e D Cannot be determined

Solution:

e C150

The next two problems refer to the following information: P(A) = 0.2,
P(B) = 0.3, A and B are independent events.
Exercise:

Problem: P(A AND B) =
e AO.5
e B0.6
e CO
e DO.06

Solution:

D

Exercise:

Problem: P(A OR B) =

¢ A0.56

e B05
e C0.44
e Dil

Solution:

Cc
Exercise:
Problem:

If H and D are mutually exclusive events, P(H) = 0.25,
P(D) = 0.15, then P(H|D)

e Al

¢e BO

e C 0.40
e D 0.0375

Solution:

B

Introduction

Student Learning Outcomes
By the end of this chapter, the student should be able to:

e Differentiate between Type I and Type II Errors

e Describe hypothesis testing in general and in practice

e Conduct and interpret hypothesis tests for a single population mean,
population standard deviation known.

¢ Conduct and interpret hypothesis tests for a single population mean,
population standard deviation unknown.

¢ Conduct and interpret hypothesis tests for a single population
proportion.

Introduction

One job of a Statistician is to make statistical inferences about populations
based on samples taken from the population. Confidence intervals are one
way to estimate a population parameter. Another way to make a Statistical
inference is to make a decision about a parameter. For instance, a car dealer
advertises that its new small truck gets 35 miles per gallon, on the average.
A tutoring service claims that its method of tutoring helps 90% of its
students get an A or a B. A company says that women managers in their
company earn an average of $60,000 per year.

A Statistician will make a decision about these claims. This process is called
"hypothesis testing." A hypothesis test involves collecting data from a
sample and evaluating the data. Then, the statistician makes a decision as to
whether or not there is sufficient evidence based upon analyses of the data,
to reject the null hypothesis.

In this chapter, you will conduct hypothesis tests on single means and single
proportions. You will also learn about the errors associated with these tests.

Hypothesis testing consists of two contradictory hypotheses or statements, a
decision based on the data, and a conclusion. To perform a hypothesis test, a

Statistician will:

1. Set up two contradictory hypotheses.

2. Collect sample data (in homework problems, the data or summary
Statistics will be given to you).

3. Determine the correct distribution to perform the hypothesis test.

4. Analyze sample data by performing the calculations that ultimately
will allow you to reject or fail to reject the null hypothesis.

5. Make a decision and write a meaningful conclusion.

Note:To do the hypothesis test homework problems for this chapter and
later chapters, make copies of the appropriate special solution sheets. See
the Table of Contents topic "Solution Sheets".

Glossary

Confidence Interval (CI)
An interval estimate for an unknown population parameter. This

depends on:

e The desired confidence level.

e Information that is known about the distribution (for example,
known standard deviation).

e The sample and its size.

Hypothesis Testing
Based on sample evidence, a procedure to determine whether the
hypothesis stated is a reasonable statement and cannot be rejected, or
is unreasonable and should be rejected.

Null and Alternate Hypotheses

The actual test begins by considering two hypotheses. They are called the
null hypothesis and the alternate hypothesis. These hypotheses contain
opposing viewpoints.

H,: The null hypothesis: It is a statement about the population that will be
assumed to be true unless it can be shown to be incorrect beyond a
reasonable doubt.

H{,: The alternate hypothesis: It is a claim about the population that is
contradictory to H, and what we conclude when we reject Ho.

Example:

H,: No more than 30% of the registered voters in Santa Clara County
voted in the primary election.

Hf,: More than 30% of the registered voters in Santa Clara County voted in
the primary election.

Example:

We want to test whether the mean grade point average in American
colleges is different from 2.0 (out of 4.0).

a ft — 2.0 lage Mise Bal)

Example:

We want to test if college students take less than five years to graduate
from college, on the average.

ed 5 0 Jib [he

Example:

In an issue of U. S. News and World Report, an article on school
standards stated that about half of all students in France, Germany, and
Israel take advanced placement exams and a third pass. The same article
stated that 6.6% of U. S. students take advanced placement exams and 4.4
% pass. Test if the percentage of U. S. students who take advanced
placement exams is more than 6.6%.

A,: p= 0.066 Ha: p > 0.066

Since the null and alternate hypotheses are contradictory, you must examine
evidence to decide if you have enough evidence to reject the null hypothesis
or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you
make a decision. There are two options for a decision. They are "reject H,"
if the sample information favors the alternate hypothesis or "do not reject
Hf," or "fail to reject H," if the sample information is insufficient to reject
the null hypothesis.

Mathematical Symbols Used in H, and A:

isk A,

- not equal (+) or greater than (>) or less
ce) than (<)
greater than or equal
to) less than (<)
less than or equal to (

<) more than (>)

Note: 1, always has a symbol with an equal in it. H, never has a symbol
with an equal in it. The choice of symbol depends on the wording of the
hypothesis test. However, be aware that many researchers (including one of
the co-authors in research work) use = in the Null Hypothesis, even with
> or < as the symbol in the Alternate Hypothesis. This practice is
acceptable because we only make the decision to reject or not reject the
Null Hypothesis.

Optional Collaborative Classroom Activity

Bring to class a newspaper, some news magazines, and some Internet
articles . In groups, find articles from which your group can write a null and
alternate hypotheses. Discuss your hypotheses with the rest of the class.

Glossary

Hypothesis
A statement about the value of a population parameter. In case of two
hypotheses, the statement assumed to be true is called the null
hypothesis (notation Ho) and the contradictory statement is called the
alternate hypothesis (notation H,).

Outcomes and the Type I and Type II Errors

When you perform a hypothesis test, there are four possible outcomes
depending on the actual truth (or falseness) of the null hypothesis H, and
the decision to reject or not. The outcomes are summarized in the following
table:

ACTION H, 1S ACTUALLY

True False
Do not reject H, Correct Outcome Type II error
Reject H, Type I Error Correct Outcome

The four possible outcomes in the table are:

e The decision is to not reject H, when, in fact, H, is true (correct
decision).

e The decision is to reject H, when, in fact, H, is true (incorrect
decision known as a Type I error).

e The decision is to not reject H, when, in fact, H, is false (incorrect
decision known as a Type I error).

e The decision is to reject H, when, in fact, H, is false (correct
decision whose probability is called the Power of the Test).

Each of the errors occurs with a particular probability. The Greek letters a
and @ represent the probabilities.

qa = probability of a Type I error = P(Type I error) = probability of
rejecting the null hypothesis when the null hypothesis is true.

@ = probability of a Type II error = P(Type II error) = probability of not
rejecting the null hypothesis when the null hypothesis is false.

a and £ should be as small as possible because they are probabilities of
errors. They are rarely 0.

The Power of the Test is 1 — @. Ideally, we want a high power that is as
close to 1 as possible. Increasing the sample size can increase the Power of
the Test.

The following are examples of Type I and Type II errors.

Example:

Suppose the null hypothesis, Ho, is: Frank's rock climbing equipment is
safe.

Type I error: Frank thinks that his rock climbing equipment may not be
safe when, in fact, it really is safe. Type II error: Frank thinks that his
rock climbing equipment may be safe when, in fact, it is not safe.

qa = probability that Frank thinks his rock climbing equipment may not be
safe when, in fact, it really is safe. 6 = probability that Frank thinks his
rock climbing equipment may be safe when, in fact, it is not safe.

Notice that, in this case, the error with the greater consequence is the Type
II error. (If Frank thinks his rock climbing equipment is safe, he will go
ahead and use it.)

Example:

Suppose the null hypothesis, H,, is: The victim of an automobile accident
is alive when he arrives at the emergency room of a hospital.

Type I error: The emergency crew thinks that the victim is dead when, in
fact, the victim is alive. Type II error: The emergency crew does not
know if the victim is alive when, in fact, the victim is dead.

qa = probability that the emergency crew thinks the victim is dead when,
in fact, he is really alive = P(Type I error). G = probability that the

emergency crew does not know if the victim is alive when, in fact, the
victim is dead = P(Type IJ error).

The error with the greater consequence is the Type I error. (If the
emergency crew thinks the victim is dead, they will not treat him.)

Glossary

Type 1 Error
The decision is to reject the Null hypothesis when, in fact, the Null

hypothesis is true.

Type 2 Error
The decision is to not reject the Null hypothesis when, in fact, the Null

hypothesis is false.

Distribution Needed for Hypothesis Testing

Earlier in the course, we discussed sampling distributions. Particular
distributions are associated with hypothesis testing. Perform tests of a
population mean using a normal distribution or a student's-t
distribution. (Remember, use a student's-t distribution when the population
standard deviation is unknown and the distribution of the sample mean is
approximately normal.) In this chapter we perform tests of a population
proportion using a normal distribution (usually n is large or the sample size
is large).

If you are testing a single population mean, the distribution for the test is
for means:

A N(ux, 3) or tar

The population parameter is jz. The estimated value (point estimate) for pu is
x, the sample mean.

If you are testing a single population proportion, the distribution for the
test is for proportions or percentages:

Pwo)

The population parameter is p. The estimated value (point estimate) for p is
p’. p’ = = where z is the number of successes and n is the sample size.

Glossary

Normal Distribution
A continuous random variable (RV) with pdf

Ex) = i e (eH)"/ 20" where yz is the mean of the distribution and

o is the standard deviation. Notation: X ~ N(y, 0). If uw = 0 and
o = 1, the RV is called the standard normal distribution.

Standard Deviation
A number that is equal to the square root of the variance and measures
how far data values are from their mean. Notation: s for sample
standard deviation and o for population standard deviation.

Student's-t Distribution
Investigated and reported by William S. Gossett in 1908 and published
under the pseudonym Student. The major characteristics of the random
variable (RV) are:

e It is continuous and assumes any real values.

e The pdf is symmetrical about its mean of zero. However, it is
more spread out and flatter at the apex than the normal
distribution.

e It approaches the standard normal distribution as n gets larger.

e There is a "family" of t distributions: every representative of the
family is completely defined by the number of degrees of
freedom which is one less than the number of data.

Assumption

When you perform a hypothesis test of a single population mean py using
a Student's-t distribution (often called a t-test), there are fundamental
assumptions that need to be met in order for the test to work properly. Your
data should be a simple random sample that comes from a population that
is approximately normally distributed. You use the sample standard
deviation to approximate the population standard deviation. (Note that if
the sample size is sufficiently large, a t-test will work even if the population
is not approximately normally distributed).

When you perform a hypothesis test of a single population mean jy using
a normal distribution (often called a z-test), you take a simple random
sample from the population. The population you are testing is normally
distributed or your sample size is sufficiently large. You know the value of
the population standard deviation.

When you perform a hypothesis test of a single population proportion p,
you take a simple random sample from the population. You must meet the
conditions for a binomial distribution which are there are a certain number
n of independent trials, the outcomes of any trial are success or failure, and
each trial has the same probability of a success p. The shape of the binomial
distribution needs to be similar to the shape of the normal distribution. To
ensure this, the quantities np and nq must both be greater than five (np > 5
and nq > 5). Then the binomial distribution of sample (estimated)
proportion can be approximated by the normal distribution with 4 = p and

o = ,/**. Remember that g = 1 — p.

Glossary

Binomial Distribution
A discrete random variable (RV) which arises from Bernoulli trials.
There are a fixed number, n, of independent trials. “Independent”
means that the result of any trial (for example, trial 1) does not affect
the results of the following trials, and all trials are conducted under the
same conditions. Under these circumstances the binomial RV X is

defined as the number of successes in 7 trials. The notation is: X~
B(n, p). The mean is js = np and the standard deviation is o = ,/npq
. The probability of exactly z successes in 7 trials is

P(X = 2) = (1 )p*q"*.

Normal Distribution
A continuous random variable (RV) with pdf

fx) = Sr e~(*-H)?/20° | where jy. is the mean of the distribution and

o is the standard deviation. Notation: X ~ N(y, 0). If uw = 0 and
ao = 1, the RV is called the standard normal distribution.

Standard Deviation
A number that is equal to the square root of the variance and measures
how far data values are from their mean. Notation: s for sample
standard deviation and o for population standard deviation.

Student-t Distribution
Investigated and reported by William S. Gossett in 1908 and published
under the pseudonym Student. The major characteristics of the random
variable (RV) are:

e It is continuous and assumes any real values.

e The pdf is symmetrical about its mean of zero. However, it is
more spread out and flatter at the apex than the normal
distribution.

e It approaches the standard normal distribution as n gets larger.

e There is a "family" of t distributions: every representative of the
family is completely defined by the number of degrees of
freedom which is one less than the number of data.

Rare Events

Suppose you make an assumption about a property of the population (this
assumption is the null hypothesis). Then you gather sample data randomly.
If the sample has properties that would be very unlikely to occur if the
assumption is true, then you would conclude that your assumption about the
population is probably incorrect. (Remember that your assumption is just an
assumption - it is not a fact and it may or may not be true. But your sample
data are real and the data are showing you a fact that seems to contradict
your assumption. )

For example, Didi and Ali are at a birthday party of a very wealthy friend.
They hurry to be first in line to grab a prize from a tall basket that they
cannot see inside because they will be blindfolded. There are 200 plastic
bubbles in the basket and Didi and Ali have been told that there is only one
with a $100 bill. Didi is the first person to reach into the basket and pull out
a bubble. Her bubble contains a $100 bill. The probability of this happening
is sr = 0.005. Because this is so unlikely, Ali is hoping that what the two

of them were told is wrong and there are more $100 bills in the basket. A
"rare event" has occurred (Didi getting the $100 bill) so Ali doubts the
assumption about only one $100 bill being in the basket.

Glossary

Hypothesis
A statement about the value of a population parameter. In case of two
hypotheses, the statement assumed to be true is called the null
hypothesis (notation Ho) and the contradictory statement is called the
alternate hypothesis (notation H,).

Using the Sample to Test the Null Hypothesis

Use the sample data to calculate the actual probability of getting the test
result, called the p-value. The p-value is the probability that, if the null
hypothesis is true, the results from another randomly selected sample
will be as extreme or more extreme as the results obtained from the
given sample.

A large p-value calculated from the data indicates that we should fail to
reject the null hypothesis. The smaller the p-value, the more unlikely the
outcome, and the stronger the evidence is against the null hypothesis. We
would reject the null hypothesis if the evidence is strongly against it.

Draw a graph that shows the p-value. The hypothesis test is easier to
perform if you use a graph because you see the problem more clearly.

Example:
(to illustrate the p-value)
Suppose a baker claims that his bread height is more than 15 cm, on the
average. Several of his customers do not believe him. To persuade his
customers that he is right, the baker decides to do a hypothesis test. He
bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm.
The baker knows from baking hundreds of loaves of bread that the
standard deviation for the height is 0.5 cm. and the distribution of heights
is normal.
The null hypothesis could be H,: ~ < 15 The alternate hypothesis is H,:
Tes
The words "is more than" translates as a'">" so "~ > 15" goes into the
alternate hypothesis. The null hypothesis must contradict the alternate
hypothesis.
Since o is known (o0 = 0.5 cm.), the distribution for the population is
known to be normal with mean = 15 and standard deviation =

0.5
Suppose the null hypothesis is true (the mean height of the loaves is no
more than 15 cm). Then is the mean height (17 cm) calculated from the

sample unexpectedly large? The hypothesis test works by asking the
question how unlikely the sample mean would be if the null hypothesis
were true. The graph shows how far out the sample mean is on the normal
curve. The p-value is the probability that, if we were to take other samples,
any other sample mean would fall at least as far out as 17 cm.

The p-value, then, is the probability that a sample mean is the same or
greater than 17 cm. when the population mean is, in fact, 15 cm. We
can calculate this probability using the normal distribution for means from
Chapter 7.

p-value is
approximately 0

15 17

p-value = P(% > 17) which is approximately 0.

A p-value of approximately 0 tells us that it is highly unlikely that a loaf of
bread rises no more than 15 cm, on the average. That is, almost 0% of all
loaves of bread would be at least as high as 17 cm. purely by CHANCE
had the population mean height really been 15 cm. Because the outcome of
17 cm. is so unlikely (meaning it is happening NOT by chance alone),
we conclude that the evidence is strongly against the null hypothesis (the
mean height is at most 15 cm.). There is sufficient evidence that the true
mean height for the population of the baker's loaves of bread is greater than
15 cm.

Glossary

Hypothesis
A statement about the value of a population parameter. In case of two
hypotheses, the statement assumed to be true is called the null
hypothesis (notation Ho) and the contradictory statement is called the
alternate hypothesis (notation H,).

p-value
The probability that an event will happen purely by chance assuming
the null hypothesis is true. The smaller the p-value, the stronger the
evidence is against the null hypothesis.

Standard Deviation
A number that is equal to the square root of the variance and measures
how far data values are from their mean. Notation: s for sample
standard deviation and o for population standard deviation.

Decision and Conclusion

A systematic way to make a decision of whether to reject or not reject the
null hypothesis is to compare the p-value and a preset or preconceived a
(also called a "significance level"). A preset a is the probability of a Type
L error (rejecting the null hypothesis when the null hypothesis is true). It
may or may not be given to you at the beginning of the problem.

When you make a decision to reject or not reject Ho, do as follows:

e If a > p-value, reject H,. The results of the sample data are
significant. There is sufficient evidence to conclude that H, is an
incorrect belief and that the alternative hypothesis, H,, may be
correct.

e If a < p-value, do not reject H,. The results of the sample data are
not significant. There is not sufficient evidence to conclude that the
alternative hypothesis, H,, may be correct.

e When you "do not reject H,", it does not mean that you should believe
that H, is true. It simply means that the sample data have failed to
provide sufficient evidence to cast serious doubt about the truthfulness
Ol ff is

Conclusion: After you make your decision, write a thoughtful conclusion
about the hypotheses in terms of the given problem.

Glossary

Hypothesis
A statement about the value of a population parameter. In case of two
hypotheses, the statement assumed to be true is called the null
hypothesis (notation Ho) and the contradictory statement is called the
alternate hypothesis (notation H,).

Level of Significance of the Test
Probability of a Type I error (reject the null hypothesis when it is true).
Notation: a. In hypothesis testing, the Level of Significance is called
the preconceived a or the preset a.

p-value
The probability that an event will happen purely by chance assuming
the null hypothesis is true. The smaller the p-value, the stronger the
evidence is against the null hypothesis.

Type 1 Error
The decision is to reject the Null hypothesis when, in fact, the Null
hypothesis is true.

Additional Information

In a hypothesis test problem, you may see words such as "the level of
significance is 1%." The "1%" is the preconceived or preset a.

The statistician setting up the hypothesis test selects the value of a@ to
use before collecting the sample data.

If no level of significance is given, the accepted standard is to use
a = 0.05.

When you calculate the p-value and draw the picture, the p-value is
the area in the left tail, the right tail, or split evenly between the two
tails. For this reason, we call the hypothesis test left, right, or two
tailed.

The alternate hypothesis, H,, tells you if the test is left, right, or two-
tailed. It is the key to conducting the appropriate test.

Hf, never has a symbol that contains an equal sign.

Thinking about the meaning of the p-value: A data analyst (and
anyone else) should have more confidence that he made the correct
decision to reject the null hypothesis with a smaller p-value (for
example, 0.001 as opposed to 0.04) even if using the 0.05 level for
alpha. Similarly, for a large p-value like 0.4, as opposed to a p-value of
0.056 (alpha = 0.05 is less than either number), a data analyst should
have more confidence that she made the correct decision in failing to
reject the null hypothesis. This makes the data analyst use judgment
rather than mindlessly applying rules.

The following examples illustrate a left, right, and two-tailed test.

Example:

Gt 0 Vela (Ve <5)

Test of a single population mean. H, tells you the test is left-tailed. The
picture of the p-value is as follows:

p-value

4 |

Example:

ig p= 0:2 ip. 0:2

This is a test of a single population proportion. H, tells you the test is
right-tailed. The picture of the p-value is as follows:

p-value

0.2

Example:

a: ff = 50 Ha? jb 50

This is a test of a single population mean. H, tells you the test is two-
tailed. The picture of the p-value is as follows.

l l
— (p-value — p-value)
5 (p-value) 5 |

4 |

50

Glossary

Hypothesis Testing
Based on sample evidence, a procedure to determine whether the
hypothesis stated is a reasonable statement and cannot be rejected, or
is unreasonable and should be rejected.

p-value
The probability that an event will happen purely by chance assuming
the null hypothesis is true. The smaller the p-value, the stronger the
evidence is against the null hypothesis.

Summary of the Hypothesis Test

The hypothesis test itself has an established process. This can be
summarized as follows:

. Determine H, and H,. Remember, they are contradictory.

. Determine the random variable.

. Determine the distribution for the test.

. Draw a graph, calculate the test statistic, and use the test statistic to
calculate the p-value. (A z-score and a t-score are examples of test
Statistics.)

5. Compare the preconceived @ with the p-value, make a decision (reject

or do not reject H,), and write a clear conclusion using English

sentences.

BRWN FP

Notice that in performing the hypothesis test, you use @ and not £. is
needed to help determine the sample size of the data that is used in
calculating the p-value. Remember that the quantity 1 — £ is called the
Power of the Test. A high power is desirable. If the power is too low,
Statisticians typically increase the sample size while keeping a the same. If
the power is low, the null hypothesis might not be rejected when it should
be.

Glossary

Hypothesis Testing
Based on sample evidence, a procedure to determine whether the
hypothesis stated is a reasonable statement and cannot be rejected, or
is unreasonable and should be rejected.

p-value
The probability that an event will happen purely by chance assuming
the null hypothesis is true. The smaller the p-value, the stronger the
evidence is against the null hypothesis.

Examples

This module provides examples of Hypothesis Testing of a Single Mean
and a Single Proportion as a part of the Collaborative Statistics collection
(col10522) by Barbara Illowsky and Susan Dean.

Example:
Exercise:

Problem:

Jeffrey, as an eight-year old, established a mean time of 16.43
seconds for swimming the 25-yard freestyle, with a standard
deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could
swim the 25-yard freestyle faster by using goggles. Frank bought
Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-
yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16
seconds. Frank thought that the goggles helped Jeffrey to swim
faster than the 16.43 seconds. Conduct a hypothesis test using a
preset a = 0.05. Assume that the swim times for the 25-yard
freestyle are normal.

Solution:
Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single
population mean.

Ho: u=16.43 Ha:w< 16.43

For Jeffrey to swim faster, his time will be less than 16.43 seconds.
The "<" tells you this is left-tailed.

Determine the distribution needed:

Random variable: X = the mean time to swim the 25-yard freestyle.

Distribution for the test: X is normal (population standard
deviation is known: o = 0.8)

ee N(n, x) Therefore, X ~ N (16.43, 8.

jt = 16.43 comes from AH and not the data. o = 0.8, andn = 15.
Calculate the p-value using the normal distribution for a mean:
p-value = P(a << 16) = 0.0187 where the sample mean in the
problem is given as 16.

p-value = 0.0187 (This is called the actual level of significance.)
The p-value is the area to the left of the sample mean is given as 16.

Graph:
p-value -
x= 16
nw = 16.43
16 16.43 ss

jt = 16.43 comes from H,. Our assumption is wy = 16.43.
Interpretation of the p-value: If H, is true, there is a 0.0187
probability (1.87%) that Jeffrey's mean time to swim the 25-yard
freestyle is 16 seconds or less. Because a 1.87% chance is small, the

mean time of 16 seconds or less is unlikely to have happened
randomly. It is a rare event.

Compare q@ and the p-value:
a = 0.05 p-value = 0.0187 a > p-value

Make a decision: Since a > p-value, reject Ho.

This means that you reject 44 = 16.43. In other words, you do not
think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster
with the new goggles.

Conclusion: At the 5% significance level, we conclude that Jeffrey
swims faster using the new goggles. The sample data show there is
sufficient evidence that Jeffrey's mean time to swim the 25-yard
freestyle is less than 16.43 seconds.

The p-value can easily be calculated using the TI-83+ and the TI-84
calculators:

Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over
to Stats and press ENTER. Arrow down and enter 16.43 for fp (null
hypothesis), .8 for 0, 16 for the sample mean, and 15 for n. Arrow
down to p: (alternate hypothesis) and arrow over to <fg. Press
ENTER. Arrow down to Calculate and press ENTER. The
calculator not only calculates the p-value (p = 0.0187) but it also
calculates the test statistic (z-score) for the sample mean. < 16.43
is the alternate hypothesis. Do this set of instructions again except
arrow to Draw (instead of Calculate). Press ENTER. A shaded
graph appears with z = —2.08 (test statistic) and p = 0.0187 (p-
value). Make sure when you use Dr aw that no other equations are
highlighted in Y = and the plots are turned off.

When the calculator does a Z-Test, the Z- Test function finds the p-
value by doing a normal probability calculation using the Central
Limit Theorem:

2 -op—2nd DISTR normcdf
(—10 WHA 16.43, 0.8/V15).
The Type I and Type IJ errors for this problem are as follows:

The Type I error is to conclude that Jeffrey swims the 25-yard
freestyle, on average, in less than 16.43 seconds when, in fact, he

actually swims the 25-yard freestyle, on average, in 16.43 seconds.
(Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey
swims the 25-yard free-style, on average, in less than 16.43 seconds
when, in fact, he actually does swim the 25-yard free-style, on
average, in less than 16.43 seconds. (Do not reject the null hypothesis
when the null hypothesis is false.)

Historical Note: The traditional way to compare the two probabilities, a
and the p-value, is to compare the critical value (z-score from q) to the test
Statistic (z-score from data). The calculated test statistic for the p-value is
—2.08. (From the Central Limit Theorem, the test statistic formula is

(=) . For this problem, x = 16, wx = 16.43 from the null
hypothesis, ox = 0.8, and m = 15.) You can find the critical value for

a = 0.05 in the normal table (see 15.Tables in the Table of Contents). The
z-score for an area to the left equal to 0.05 is midway between -1.65 and
-1.64 (0.05 is midway between 0.0505 and 0.0495). The z-score is -1.645.
Since —1.645 > —2.08 (which demonstrates that @ > p-value), reject Ho.
Traditionally, the decision to reject or not reject was done in this way.
Today, comparing the two probabilities a and the p-value is very common.
For this problem, the p-value, 0.0187 is considerably smaller than a, 0.05.
You can be confident about your decision to reject. The graph shows a, the
p-value, and the test statistics and the critical value.

—

p-value = 0.0187

-2.08 -1.645 0

Example:
Exercise:

Problem:

A college football coach thought that his players could bench press a
mean weight of 275 pounds. It is known that the standard deviation
is 55 pounds. Three of his players thought that the mean weight was
more than that amount. They asked 30 of their teammates for their
estimated maximum lift on the bench press exercise. The data ranged
from 205 pounds to 385 pounds. The actual different weights were
(frequencies are in parentheses) 205(3) 215(3) 225(1) 241(2) 252(2)
265(2) 275(2) 313(2) 316(5) 338(2) 341(1) 345(2) 368(2) 385(1).
(Source: data from Reuben Davis, Kraig Evans, and Scott
Gunderson. )

Conduct a hypothesis test using a 2.5% level of significance to
determine if the bench press mean is more than 275 pounds.

Solution:
Set up the Hypothesis Test:

Since the problem is about a mean weight, this is a test of a single
population mean.

Hi — 2 J aT eT hS) This is a right-tailed test.
Calculating the distribution needed:

Random variable: X = the mean weight, in pounds, lifted by the
football players.

Distribution for the test: It is normal because o is known.

z 55
Kaen) (275, $5.)

x = 286.2 pounds (from the data).

o = 55 pounds (Always use o if you know it.) We assume fz = 275
pounds unless our data shows us otherwise.

Calculate the p-value using the normal distribution for a mean and
using the sample mean as input (see the calculator instructions below
for using the data as input):

p-value = P( x > 286.2) = 0.1323.

Interpretation of the p-value: If H, is true, then there is a 0.1331
probability (13.23%) that the football players can lift a mean weight
of 286.2 pounds or more. Because a 13.23% chance is large enough, a
mean weight lift of 286.2 pounds or more is not a rare event.

x= 286.2 p-value = 0.1323
w= 275

275 286.2 x

Compare a and the p-value:
w= 0025 p-value = 0.1323
Make a decision: Since a<p-value, do not reject H,.

Conclusion: At the 2.5% level of significance, from the sample data,
there is not sufficient evidence to conclude that the true mean weight
lifted is more than 275 pounds.

The p-value can easily be calculated using the TI-83+ and the TI-84
calculators:

Put the data and frequencies into lists. Press STAT and arrow over to
TESTS. Press 1:Z-Test. Arrow over to Data and press ENTER.

Arrow down and enter 275 for 49, 55 for a, the name of the list where
you put the data, and the name of the list where you put the
frequencies. Arrow down to p : and arrow over to > [Up. Press
ENTER. Arrow down to Calculate and press ENTER. The
calculator not only calculates the p-value (p = 0.1331, a little
different from the above calculation - in it we used the sample mean
rounded to one decimal place instead of the data) but it also calculates
the test statistic (z-score) for the sample mean, the sample mean, and
the sample standard deviation. 4 > 275 is the alternate hypothesis.
Do this set of instructions again except arrow to Dr aw (instead of
Calculate). Press ENTER. A shaded graph appears with z = 1.112
(test statistic) and p = 0.1331 (p-value). Make sure when you use
Draw that no other equations are highlighted in Y = and the plots are
turned off.

Example:
Exercise:

Problem:

Statistics students believe that the mean score on the first statistics test
is 65. A statistics instructor thinks the mean score is higher than 65.
He samples ten statistics students and obtains the scores 65 65 70 67
66 63 63 68 72 71. He performs a hypothesis test using a 5% level of
significance. The data are from a normal distribution.

Solution:
Set up the Hypothesis Test:

A 5% level of significance means that a = 0.05. This is a test of a
single population mean.

Heo — op aie = 65

Since the instructor thinks the average score is higher, use a ">". The
">" means the test is right-tailed.

Determine the distribution needed:

Random variable: X = average score on the first statistics test.
Distribution for the test: If you read the problem carefully, you will
notice that there is no population standard deviation given. You are
only given n = 10 sample data values. Notice also that the data come
from a normal distribution. This means that the distribution for the

test is a student's-t.

Use tar. Therefore, the distribution for the test is tj) where n = 10 and
df =10—-1=9.

Calculate the p-value using the Student's-t distribution:

p-value = P( x > 67 )= 0.0396 where the sample mean and sample
standard deviation are calculated as 67 and 3.1972 from the data.

Interpretation of the p-value: If the null hypothesis is true, then

there is a 0.0396 probability (3.96%) that the sample mean is 67 or
more.

p-value = 0.0396

65 67 *

Compare q@ and the p-value:

Since a = .05 and p-value = 0.0396. Therefore, a > p-value.

Make a decision: Since a > p-value, reject Ho.

This means you reject p42 = 65. In other words, you believe the
average test score is more than 65.

Conclusion: At a 5% level of significance, the sample data show
sufficient evidence that the mean (average) test score is more than 65,
just as the math instructor thinks.

The p-value can easily be calculated using the TI-83+ and the TI-84
calculators:

Put the data into a list. Press STAT and arrow over to TESTS. Press
2:T-Test. Arrow over to Data and press ENTER. Arrow down and
enter 65 for jz, the name of the list where you put the data, and 1 for
Freq:. Arrow down to p : and arrow over to > fo. Press ENTER.
Arrow down to Calculate and press ENTER. The calculator not
only calculates the p-value (p = 0.0396) but it also calculates the test
Statistic (t-score) for the sample mean, the sample mean, and the
sample standard deviation. jz > 65 is the alternate hypothesis. Do this
set of instructions again except arrow to Dr aw (instead of
Calculate). Press ENTER. A shaded graph appears with

t = 1.9781 (test statistic) and p = 0.0396 (p-value). Make sure when
you use Draw that no other equations are highlighted in Y = and the
plots are turned off.

Example:
Exercise:

Problem:

Joon believes that 50% of first-time brides in the United States are
younger than their grooms. She performs a hypothesis test to
determine if the percentage is the same or different from 50%. Joon
samples 100 first-time brides and 53 reply that they are younger than
their grooms. For the hypothesis test, she uses a 1% level of
significance.

Solution:
Set up the Hypothesis Test:

The 1% level of significance means that a = 0.01. This is a test of a
single population proportion.

Hep ol) in ipra oe = 0p a9)

The words "is the same or different from" tell you this is a two-
tailed test.

Calculate the distribution needed:

Random variable: P/ = the percent of of first-time brides who are
younger than their grooms.

Distribution for the test: The problem contains no mention of a

mean. The information is given in terms of percentages. Use the
distribution for P/, the estimated proportion.

Pi~N (», 4/ = Therefore, Pr ~ N (0, ae where
p— 0: 50eg— bp 0 50 anda — 100;
Calculate the p-value using the normal distribution for proportions:

p-value = P(p’< 0.47 or p’ > 0.53 ) = 0.5485

rae ee Se
where £ = 53, pl= — = ito = 0.53.

Interpretation of the p-value: If the null hypothesis is true, there is
0.5485 probability (54.85%) that the sample (estimated) proportion p/
is 0.53 or more OR 0.47 or less (see the graph below).

= (p-value) = 0.27425 = (p-value) = 0.27425

0.47 0.50 0.53

jt = p = 0.50 comes from A, the null hypothesis.

pl= 0.53. Since the curve is symmetrical and the test is two-tailed,
the p/ for the left tail is equal to 0.50 — 0.03 = 0.47 where
pL = p = 0.50. (0.03 is the difference between 0.53 and 0.50.)

Compare q@ and the p-value:
Since a = 0.01 and p-value = 0.5485. Therefore, a< p-value.
Make a decision: Since a<p-value, you cannot reject Ho.

Conclusion: At the 1% level of significance, the sample data do not
show sufficient evidence that the percentage of first-time brides that
are younger than their grooms is different from 50%.

The p-value can easily be calculated using the TI-83+ and the TI-84
calculators:

Press STAT and arrow over to TESTS. Press 5:1-PropZTest.
Enter .5 for po, 53 for x and 100 for n. Arrow down to Prop and
arrow tonot equals po. Press ENTER. Arrow down to

Calculate and press ENTER. The calculator calculates the p-value
(p = 0.5485) and the test statistic (z-score). Prop not equals .5
is the alternate hypothesis. Do this set of instructions again except
arrow to Draw (instead of Calculate). Press ENTER. A shaded
graph appears with z = 0.6 (test statistic) and p = 0.5485 (p-value).
Make sure when you use Dr aw that no other equations are
highlighted in Y = and the plots are turned off.

The Type I and Type IJ errors are as follows:

The Type I error is to conclude that the proportion of first-time brides
that are younger than their grooms is different from 50% when, in
fact, the proportion is actually 50%. (Reject the null hypothesis when
the null hypothesis is true).

The Type II error is there is not enough evidence to conclude that the
proportion of first time brides that are younger than their grooms
differs from 50% when, in fact, the proportion does differ from 50%.
(Do not reject the null hypothesis when the null hypothesis is false.)

Example:
Exercise:

Problem:

Suppose a consumer group suspects that the proportion of households
that have three cell phones is 30%. A cell phone company has reason
to believe that the proportion is 30%. Before they start a big
advertising campaign, they conduct a hypothesis test. Their marketing
people survey 150 households with the result that 43 of the
households have three cell phones.

Solution:

Set up the Hypothesis Test:

Hp ol) ipo)
Determine the distribution needed:

The random variable is P/ = proportion of households that have
three cell phones.

The distribution for the hypothesis test is P/ ~ N
(0.30)-(0.70)
(0.30, ee |

Exercise:

Problem:
The value that helps determine the p-value is p/. Calculate p/.

Solution:

pl= = where zx is the number of successes and 7 is the

total number in the sample.

L—=—Ads =o)
es
P50)
Exercise:

Problem: What is a success for this problem?

Solution:

A success is having three cell phones in a household.

Exercise:

Problem: What is the level of significance?

Solution:

The level of significance is the preset a. Since a is not given,
assume that a = 0.05.

Draw the graph for this problem. Draw the horizontal axis. Label and
shade appropriately.
Exercise:

Problem: Calculate the p-value.

Solution:

p-value = 0.7216
Exercise:

Problem:

Make a decision. (Reject/Do not reject) Ho
because

Solution:

Assuming that a = 0.05, a < p-value. The Decision is do not
reject Ho because there is not sufficient evidence to conclude
that the proportion of households that have three cell phones is
not 30%.

The next example is a poem written by a statistics student named Nicole
Hart. The solution to the problem follows the poem. Notice that the
hypothesis test is for a single population proportion. This means that the
null and alternate hypotheses use the parameter p. The distribution for the
test is normal. The estimated proportion p/ is the proportion of fleas killed
to the total fleas found on Fido. This is sample information. The problem
gives a preconceived a = 0.01, for comparison, and a 95% confidence
interval computation. The poem is clever and humorous, so please enjoy it!

Note:Hypothesis testing problems consist of multiple steps. To help you do
the problems, solution sheets are provided for your use. Look in the Table
of Contents Appendix for the topic "Solution Sheets." If you like, use
copies of the appropriate solution sheet for homework problems.

Example:
Exercise:

Problem:

My dog has so many fleas, They do not come off
with ease. As for shampoo, I have tried many
types Even one called Bubble Hype, Which only
killed 25% of the fleas, Unfortunately I was
not pleased. I've used all kinds of soap, Until
I had give up hope Until one day I saw An ad
that put me in awe. A shampoo used for dogs
Called GOOD ENOUGH to Clean a Hog Guaranteed to
Kill more fleas. I gave Fido a bath And after
doing the math His number of fleas Started
dropping by 3's! Before his shampoo I counted
42. At the end of his bath, I redid the math
And the new shampoo had killed 17 fleas. So now
I was pleased. Now it is time for you to have
some fun With the level of significance being
.01, You must help me figure out Use the new
shampoo or go without?

Solution:
Set up the Hypothesis Test:
Jali — Ue ep = 0.25

Determine the distribution needed:

In words, CLEARLY state what your random variable X or P’
represents.

P’ = The proportion of fleas that are killed by the new shampoo

State the distribution to use for the test.
Normal: w (0.25, je )

Test Statistic: z = 2.3163
Calculate the p-value using the normal distribution for proportions:
p-value =0.0103

In 1 — 2 complete sentences, explain what the p-value means for this
problem.

If the null hypothesis is true (the proportion is 0.25), then there is a
0.0103 probability that the sample (estimated) proportion is 0.4048
(+) or more.

Use the previous information to sketch a picture of this situation.
CLEARLY, label and scale the horizontal axis and shade the region(s)
corresponding to the p-value.

p'
0.25 17/42= _ test statistic for
0.4048 17/42: 2.3163

Compare q@ and the p-value:

Indicate the correct decision (“reject” or “do not reject” the null
hypothesis), the reason for it, and write an appropriate conclusion,

using COMPERTE SENTENCES:

alpha decision reason for decision

0.01 Do not reject H, a<p-value

Conclusion: At the 1% level of significance, the sample data do not
show sufficient evidence that the percentage of fleas that are killed by
the new shampoo is more than 25%.

Construct a 95% Confidence Interval for the true mean or proportion.
Include a sketch of the graph of the situation. Label the point estimate
and the lower and upper bounds of the Confidence Interval.

p'

0.26 17/42 0.55

Confidence Interval: (0.26, 0.55) We are 95% confident that the true
population proportion p of fleas that are killed by the new shampoo is
between 26% and 55%.

Note: This test result is not very definitive since the p-value is very
close to alpha. In reality, one would probably do more tests by giving
the dog another bath after the fleas have had a chance to return.

Glossary

Central Limit Theorem
Given a random variable (RV) with known mean p and known
standard deviation o. We are sampling with size n and we are
interested in two new RVs - the sample mean, X, and the sample sum,
3/X. If the size n of the sample is sufficiently large, then X ~

N(n, =) and XX ~ N(nu, no). If the size n of the sample is

sufficiently large, then the distribution of the sample means and the
distribution of the sample sums will approximate a normal distribution
regardless of the shape of the population. The mean of the sample
means will equal the population mean and the mean of the sample
sums will equal n times the population mean. The standard deviation
of the distribution of the sample means, Te is called the standard

error of the mean.

Standard Deviation
A number that is equal to the square root of the variance and measures
how far data values are from their mean. Notation: s for sample
standard deviation and o for population standard deviation.

Summary of Formulas

A, and H, are contradictory.

If greater than less than
H; equal or equal to or equal to
has:

hen not equal or greater
sag d less than greater
Hi, than or less than

than

has:

Ifa p-value, then do not reject Ho.
Ifa p-value, then reject H, .

a is preconceived. Its value is set before the hypothesis test starts. The p-
value is calculated from the data.

a = probability of a Type I error = P(Type I error) = probability of rejecting
the null hypothesis when the null hypothesis is true.

@ = probability of a Type II error = P(Type II error) = probability of not
rejecting the null hypothesis when the null hypothesis is false.

If there is no given preconceived a, then use a
Types of Hypothesis Tests

e Single population mean, known population variance (or standard
deviation): Normal test.

e Single population mean, unknown population variance (or standard
deviation): Student's-t test.

e Single population proportion: Normal test.

Practice 1: Single Mean, Known Population Standard Deviation

This module provides a practice of Hypothesis Testing of Single Mean and
Single Proportion as a part of Collaborative Statistics collection (col10522)
by Barbara Illowsky and Susan Dean.

Student Learning Outcomes

e The student will conduct a hypothesis test of a single mean with
known population standard deviation.

Given

Suppose that a recent article stated that the mean time spent in jail by a
first-time convicted burglar is 2.5 years. A study was then done to see if the
mean time has increased in the new century. A random sample of 26 first—
time convicted burglars in a recent year was picked. The mean length of
time in jail from the survey was 3 years with a standard deviation of 1.8
years. Suppose that it is somehow known that the population standard
deviation is 1.5. Conduct a hypothesis test to determine if the mean length
of jail time has increased. The distribution of the population is normal.

Hypothesis Testing: Single Mean

Exercise:

Problem: Is this a test of means or proportions?
Solution:

Means

Exercise:

Problem: State the null and alternative hypotheses.

e aH:

¢ b d.::

Solution:

e a Hoi = 2.5 (or, Ho:p < 2.5)
© by: > 2.5

Exercise:
Problem:

Is this a right-tailed, left-tailed, or two-tailed test? How do you know?

Solution:
right-tailed

Exercise:

Problem: What symbol represents the Random Variable for this test?

Solution:
xX

Exercise:

Problem: In words, define the Random Variable for this test.

Solution:

The mean time spent in jail for 26 first time convicted burglars
Exercise:

Problem:

Is the population standard deviation known and, if so, what is it?

Solution:
Yes, 1.5
Exercise:
Problem:Calculate the following:
eaxrt=—
e bo=
ec Sx =
e d t=
Solution:
e a3
e b1.5

e c1.8
e d26

Exercise:

Problem:

Since both o and s, are given, which should be used? In 1 -2 complete
sentences, explain why.

Solution:

oO

Exercise:

Problem: State the distribution to use for the hypothesis test.

Solution:

7 ‘ASL
X-N(2.5,15)

Exercise:
Problem:
Sketch a graph of the situation. Label the horizontal axis. Mark the

hypothesized mean and the sample mean x. Shade the area
corresponding to the p-value.

=
x

Exercise:

Problem: Find the p-value.
Solution:
0.0446
Exercise:
Problem: At a pre-conceived a@ = 0.05, what is your:

e a Decision:
e b Reason for the decision:
e c Conclusion (write out in a complete sentence):

Solution:

e aReject the null hypothesis

Discussion Questions

Exercise:

Problem:

Does it appear that the mean jail time spent for first time convicted
burglars has increased? Why or why not?

Practice 2: Single Mean, Unknown Population Standard Deviation

This module provides a practice of Hypothesis Testing of Single Mean and
Single Proportion as a part of Collaborative Statistics collection (col10522)
by Barbara Illowsky and Susan Dean.

Student Learning Outcomes

e The student will conduct a hypothesis test of a single mean with
unknown population standard deviation.

Given
A random survey of 75 death row inmates revealed that the mean length of
time on death row is 17.4 years with a standard deviation of 6.3 years.

Conduct a hypothesis test to determine if the population mean time on death
row could likely be 15 years.

Hypothesis Testing: Single Mean

Exercise:

Problem: Is this a test of means or proportions?
Solution:
averages

Exercise:

Problem: State the null and alternative hypotheses.

e aZHu:
e bdH,:

Solution:

© ao:yw=15
© bA yw #15

Exercise:
Problem:
Is this a right-tailed, left-tailed, or two-tailed test? How do you know?
Solution:
two-tailed

Exercise:

Problem: What symbol represents the Random Variable for this test?
Solution:
Ȣ

Exercise:

Problem: In words, define the Random Variable for this test.

Solution:

the mean time spent on death row for the 75 inmates
Exercise:

Problem:
Is the population standard deviation known and, if so, what is it?
Solution:

No

Exercise:

Problem: Calculate the following:
eaxrt=—
© b6.3 =
e cn =
Solution:
e al74

e bs
e c/5

Exercise:
Problem:
Which test should be used? In 1 -2 complete sentences, explain why.
Solution:
t —test

Exercise:

Problem: State the distribution to use for the hypothesis test.
Solution:

t74
Exercise:
Problem:
Sketch a graph of the situation. Label the horizontal axis. Mark the

hypothesized mean and the sample mean, zx. Shade the area
corresponding to the p-value.

ee
x
Exercise:

Problem: Find the p-value.

Solution:
0.0015
Exercise:

Problem: At a pre-conceived a@ = 0.05, what is your:
e a Decision:
e b Reason for the decision:
e c Conclusion (write out in a complete sentence):

Solution:

e aReject the null hypothesis

Discussion Question

Does it appear that the mean time on death row could be 15 years? Why or
why not?

Practice 3: Single Proportion

This module provides a practice of Hypothesis Testing of Single Mean and
Single Proportion as a part of Collaborative Statistics collection (col10522)
by Barbara Illowsky and Susan Dean.

Student Learning Outcomes

e The student will conduct a hypothesis test of a single population
proportion.

Given

The National Institute of Mental Health published an article stating that in
any one-year period, approximately 9.5 percent of American adults suffer
from depression or a depressive illness.
(http://www.nimh.nih.gov/publicat/depression.cfm) Suppose that in a
survey of 100 people in a certain town, seven of them suffered from
depression or a depressive illness. Conduct a hypothesis test to determine if
the true proportion of people in that town suffering from depression or a
depressive illness is lower than the percent in the general adult American
population.

Hypothesis Testing: Single Proportion

Exercise:

Problem: Is this a test of means or proportions?

Solution:

Proportions

Exercise:

Problem: State the null and alternative hypotheses.

e aH:
e bd:
Solution:

© a Ho:p = 0.095
© b Aa:p < 0.095

Exercise:
Problem:
Is this a right-tailed, left-tailed, or two-tailed test? How do you know?
Solution:
left-tailed

Exercise:

Problem: What symbol represents the Random Variable for this test?

Solution:
Pr

Exercise:

Problem: In words, define the Random Variable for this test.

Solution:

the proportion of people in that town surveyed suffering from
depression or a depressive illness

Exercise:

Problem:Calculate the following:

eat=—
e bn =
e Cc p/=

Solution:
e a/

e b100
e c0.07

Exercise:
Problem:

Calculate o,,. Make sure to show how you set up the formula.

Solution:

0.0293

Exercise:

Problem: State the distribution to use for the hypothesis test.

Solution:

Normal
Exercise:
Problem:
Sketch a graph of the situation. Label the horizontal axis. Mark the

hypothesized mean and the sample proportion, p-hat. Shade the area
corresponding to the p-value.

P'
Exercise:

Problem: Find the p-value
Solution:
0.1969
Exercise:
Problem: At a pre-conceived a = 0.05, what is your:

e a Decision:
¢ b Reason for the decision:
¢ c Conclusion (write out in a complete sentence):

Solution:

e aDo not reject the null hypothesis

Discusion Question

Exercise:
Problem:
Does it appear that the proportion of people in that town with

depression or a depressive illness is lower than general adult American
population? Why or why not?

Homework

This module provides a homework of Hypothesis Testing of Single Mean
and Single Proportion as a part of Collaborative Statistics collection
(col10522) by Barbara Illowsky and Susan Dean.

Exercise:

Problem:

Some of the statements below refer to the null hypothesis, some to the
alternate hypothesis.

State the null hypothesis, H,, and the alternative hypothesis, Hg, in
terms of the appropriate parameter (jz or p).

aThe mean number of years Americans work before retiring is 34.
bAt most 60% of Americans vote in presidential elections.

cThe mean starting salary for San Jose State University graduates
is at least $100,000 per year.

d29% of high school seniors get drunk each month.

eFewer than 5% of adults ride the bus to work in Los Angeles.
fThe mean number of cars a person owns in her lifetime is not
more than 10.

gAbout half of Americans prefer to live away from cities, given
the choice.

hEuropeans have a mean paid vacation each year of six weeks.
iThe chance of developing breast cancer is under 11% for women.
jPrivate universities mean tuition cost is more than $20,000 per
year.

Solution:

a Ho: = 34; Ay:p 4 34

c H,:u > 100,000 ; H,:u < 100,000
dA y= 0.29 Aap 0.29

oH p= 0.50; Hep + 0:50

LH OAL hp O11

Exercise:

Problem:

For (a) - (j) above, state the Type I and Type II errors in complete
sentences.

Solution:

e alype I error: We conclude that the mean is not 34 years, when it
really is 34 years. Type II error: We do not conclude that the mean
is not 34 years, when it is not really 34 years.

¢ clype I error: We conclude that the mean is less than $100,000,
when it really is at least $100,000. Type II error: We do not
conclude that the mean is less than $100,000, when it is really
less than $100,000.

e dType I error: We conclude that the proportion of h.s. seniors who
get drunk each month is not 29%, when it really is 29%. Type II
error: We do not conclude that the proportion of h.s. seniors that
get drunk each month is not 29%, when it is really not 29%.

e iType I error: We conclude that the proportion is less than 11%,
when it is really at least 11%. Type II error: We do not conclude
that the proportion is less than 11%, when it really is less than
11%.

Exercise:

Problem: For (a) - (j) above, in complete sentences:

e aState a consequence of committing a Type I error.
e bState a consequence of committing a Type II error.

Note:For each of the word problems, use a solution sheet to do the
hypothesis test. The solution sheet is found in 14. Appendix (online book

version: the link is "Solution Sheets"; PDF book version: look under 14.5
Solution Sheets). Please feel free to make copies of the solution sheets. For
the online version of the book, it is suggested that you copy the .doc or the
.pdf files.

Note:If you are using a student's-t distribution for a homework problem
below, you may assume that the underlying population is normally
distributed. (In general, you must first prove that assumption, though.)

Exercise:

Problem:

A particular brand of tires claims that its deluxe tire averages at least
50,000 miles before it needs to be replaced. From past studies of this
tire, the standard deviation is known to be 8000. A survey of owners of
that tire design is conducted. From the 28 tires surveyed, the mean
lifespan was 46,500 miles with a standard deviation of 9800 miles. Do
the data support the claim at the 5% level?

Exercise:

Problem:

From generation to generation, the mean age when smokers first start
to smoke varies. However, the standard deviation of that age remains
constant of around 2.1 years. A survey of 40 smokers of this
generation was done to see if the mean starting age is at least 19. The
sample mean was 18.1 with a sample standard deviation of 1.3. Do the
data support the claim at the 5% level?

Solution:

eez— —2.71
e £0.0034
e hDecision: Reject null; Conclusion: uw < 19

e i (17.449,18.757)

Exercise:

Problem:

The cost of a daily newspaper varies from city to city. However, the
variation among prices remains steady with a standard deviation of
20¢. A study was done to test the claim that the mean cost of a daily
newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a
standard deviation of 18¢. Do the data support the claim at the 1%
level?

Exercise:

Problem:

An article in the San Jose Mercury News stated that students in the
California state university system take 4.5 years, on average, to finish
their undergraduate degrees. Suppose you believe that the mean time is
longer. You conduct a survey of 49 students and obtain a sample mean
of 5.1 with a sample standard deviation of 1.2. Do the data support
your claim at the 1% level?

Solution:

° e3.5

¢ f0.0005

e hDecision: Reject null; Conclusion: > 4.5
e i (4.7553,5.4447)

Exercise:

Problem:

The mean number of sick days an employee takes per year is believed
to be about 10. Members of a personnel department do not believe this
figure. They randomly survey 8 employees. The number of sick days
they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let
x = the number of sick days they took for the past year. Should the
personnel team believe that the mean number is about 10?

Exercise:

Problem:

In 1955, Life Magazine reported that the 25 year-old mother of three
worked, on average, an 80 hour week. Recently, many groups have
been studying whether or not the women's movement has, in fact,
resulted in an increase in the average work week for women
(combining employment and at-home work). Suppose a study was
done to determine if the mean work week has increased. 81 women
were surveyed with the following results. The sample mean was 83;
the sample standard deviation was 10. Does it appear that the mean
work week has increased for women at the 5% level?

Solution:

e e2.7

e £0.0042

e hDecision: Reject Null
e i (80.789,85.211)

Exercise:

Problem:

Your statistics instructor claims that 60 percent of the students who
take her Elementary Statistics class go through life feeling more
enriched. For some reason that she can't quite figure out, most people
don't believe her. You decide to check this out on your own. You
randomly survey 64 of her past Elementary Statistics students and find
that 34 feel more enriched as a result of her class. Now, what do you
think?

Exercise:

Problem:

A Nissan Motor Corporation advertisement read, “The average man’s
I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man
catch brown trout?” Suppose you believe that the brown trout’s mean
I.Q. is greater than 4. You catch 12 brown trout. A fish psychologist
determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5.
Conduct a hypothesis test of your belief.

Solution:

e dti

e e1.96

e £0.0380

e hDecision: Reject null when a = 0.05 ; do not reject null when
a = 0.01

e i (3.8865,5.9468)

Exercise:
Problem:
Refer to the previous problem. Conduct a hypothesis test to see if your

decision and conclusion would change if your belief were that the
brown trout’s mean I.Q. is not 4.

Exercise:

Problem:

According to an article in Newsweek, the natural ratio of girls to boys
is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose
you don’t believe the reported figures of the percent of girls born in
China. You conduct a study. In this study, you count the number of
girls and boys born in 150 randomly chosen recent births. There are 60
girls and 90 boys born of the 150. Based on your study, do you believe
that the percent of girls born in China is 46.7?

Solution:

e e-1.64
e £0.1000

e hDecision: Do not reject null
e i (0.3216,0.4784)

Exercise:

Problem:

A poll done for Newsweek found that 13% of Americans have seen or
sensed the presence of an angel. A contingent doubts that the percent is
really that high. It conducts its own survey. Out of 76 Americans
surveyed, only 2 had seen or sensed the presence of an angel. As a
result of the contingent’s survey, would you agree with the Newsweek
poll? In complete sentences, also give three reasons why the two polls
might give different results.

Exercise:

Problem:

The mean work week for engineers in a start-up company is believed
to be about 60 hours. A newly hired engineer hopes that it’s shorter.
She asks 10 engineering friends in start-ups for the lengths of their
mean work weeks. Based on the results that follow, should she count
on the mean work week to be shorter than 60 hours?

Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50;
aoe

Solution:

id dto

e e-1.33

e £0.1086

e hDecision: Do not reject null
e i (51.886,62.114)

Exercise:
Problem:
Use the “Lap time” data for Lap 4 (see Table of Contents) to test the

claim that Terri finishes Lap 4, on average, in less than 129 seconds.
Use all twenty races given.

Exercise:
Problem:
Use the “Initial Public Offering” data (see Table of Contents) to test
the claim that the mean offer price was $18 per share. Do not use all

the data. Use your random number generator to randomly survey 15
prices.

Note:The following questions were written by past students. They are
excellent problems!

Exercise:

Problem:18. "Asian Family Reunion" by Chau Nguyen

Every two years it comes around

We all get together from different towns.
In my honest opinion

It's not a typical family reunion

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout
One minute they're happy, another they'll pout.
The teenagers would look, stare, and compare
From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.
The women get tired from all of the chats
They head to the kitchen to set out the mats.
Some would sit and some would stand

Eating and talking with plates in their hands.
Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

She said that boys and men will always have
their pride,

So we won't ever see them cry.

I myself don't think she's correct,

So could you please try this problem to see if
you object?

Exercise:

Problem:"The Problem with Angels" by Cyndy Dowling

Although this problem is wholly mine,

The catalyst came from the magazine, Time.
On the magazine cover I did find

The realm of angels tickling my mind.

Inside, 69% I found to be
In angels, Americans do believe.

Then, it was time to rise to the task,
Ninety-five high school and college students I
did ask.

Viewing all as one group,

Random sampling to get the scoop.

So, I asked each to be true,
"Do you believe in angels?" Tell me, do!

Hypothesizing at the start,
Totally believing in my heart
That the proportion who said yes
Would be equal on this test.

Lo and behold, seventy-three did arrive,
Out of the sample of ninety-five.

Now your job has just begun,

Solve this problem and have some fun.

Solution:

e e1.65

e £0.0984

e hDecision: Do not reject null
e i (0.6836,0.8533)

Exercise:

Problem:"Blowing Bubbles" by Sondra Prull

Studying stats just made me tense,
I had to find some sane defense.
Some light and lifting simple play
To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress
Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least
22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows
Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax
But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

Exercise:

Problem:21. "Dalmatian Damation" by Kathy Sparling

A greedy dog breeder named Spreckles
Bred puppies with numerous freckles
The Dalmatians he sought

Possessed spot upon spot
The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.
They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove
This strategy was a wrong move.
Breeding only for spots

Would wreak havoc, they thought.
His theory they want to disprove.

They proposed a contest to Spreckles
Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.
“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one
Dalmatians that fetched a good sum.
They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

Solution:

e e-2.39

e £0.0093

e hDecision: Reject null
¢ i (91.854,100.15)

Exercise:

Problem:

"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle
Hall

As a poor starving student I don't have much money to spend for even
the bare necessities. So my favorite and main staple food is macaroni
and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious
craving for this, oh, so important, food of my life. So I went down the
street to Greatway to get a box of macaroni and cheese, but it was SO
expensive! $2.02 !!! Can you believe it? It made me stop and think.
The world is changing fast. I had thought that the mean cost of a box
(the normal size, not some super-gigantic-family-value-pack) was at
most $1, but now I wasn't so sure. However, I was determined to find
out. I went to 53 of the closest grocery stores and surveyed the prices
of macaroni and cheese. Here are the data I wrote in my notebook:
Price per box of Mac and Cheese:

e 5 stores @ $2.02
e 15 stores @ $0.25

e 3stores @ $1.29
e 6 stores @ $0.35
e Astores @ $2.27
e 7 stores @ $1.50
e 5 stores @ $1.89
e 8 stores @ 0.75.

I could see that the costs varied but I had to sit down to figure out
whether or not I was right. If it does turn out that this mouth-watering
dish is at most $1, then I'll throw a big cheesy party in our next
statistics lab, with enough macaroni and cheese for just me. (After all,
as a poor starving student I can't be expected to feed our class of
animals!)

Exercise:
Problem:
"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark"

by Jacqueline Ghodsi
THE CHARACTERS (in order of appearance):

¢ HAMLET, Prince of Denmark and student of Statistics
e POLONIUS, Hamlet’s tutor
e HOROTIO, friend to Hamlet and fellow student

Scene: The great library of the castle, in which Hamlet does his lessons
Act I

(The day is fair, but the face of Hamlet is clouded. He paces the large
room. His tutor, Polonius, is reprimanding Hamlet regarding the
latter’s recent experience. Horatio is seated at the large table at right
stage.)

POLONIUS: My Lord, how cans’t thou admit that thou hast seen a
ghost! It is but a figment of your imagination!

HAMLET: I beg to differ; I know of a certainty that five-and-seventy
in one hundred of us, condemned to the whips and scorns of time as
we are, have gazed upon a spirit of health, or goblin damn’d, be their
intents wicked or charitable.

POLONIUS If thou doest insist upon thy wretched vision then let me
invest your time; be true to thy work and speak to me through the
reason of the null and alternate hypotheses. (He turns to Horatio.) Did
not Hamlet himself say, “What piece of work is man, how noble in
reason, how infinite in faculties? Then let not this foolishness persist.
Go, Horatio, make a survey of three-and-sixty and discover what the
true proportion be. For my part, I will never succumb to this fantasy,
but deem man to be devoid of all reason should thy proposal of at least
five-and-seventy in one hundred hold true.

HORATIO (to Hamlet): What should we do, my Lord?
HAMLET: Go to thy purpose, Horatio.
HORATIO: To what end, my Lord?

HAMLET: That you must teach me. But let me conjure you by the
rights of our fellowship, by the consonance of our youth, but the
obligation of our ever-preserved love, be even and direct with me,
whether I am right or no.

(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)
Act Il

(The next day, Hamlet awaits anxiously the presence of his friend,
Horatio. Polonius enters and places some books upon the table just a
moment before Horatio enters.)

POLONIUS: So, Horatio, what is it thou didst reveal through thy
deliberations?

HORATIO: In a random survey, for which purpose thou thyself sent
me forth, I did discover that one-and-forty believe fervently that the
spirits of the dead walk with us. Before my God, I might not this
believe, without the sensible and true avouch of mine own eyes.

POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius
turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio,
let us go together, for this is not our test. (Horatio and Polonius leave
together.)

HAMLET: To reject, or not reject, that is the question: whether ‘tis
nobler in the mind to suffer the slings and arrows of outrageous
Statistics, or to take arms against a sea of data, and, by opposing, end
them. (Hamlet resignedly attends to his task.)

(Curtain falls)

Solution:

e e-1.82
e (0.0345

e hDecision: Do not reject null
e i (0.5331,0.7685)

Exercise:

Problem:"Untitled" by Stephen Chen

I've often wondered how software is released and sold to the public.
Ironically, I work for a company that sells products with known
problems. Unfortunately, most of the problems are difficult to create,
which makes them difficult to fix. I usually use the test program X,
which tests the product, to try to create a specific problem. When the
test program is run to make an error occur, the likelihood of generating
an error is 1%.

So, armed with this knowledge, I wrote a new test program Y that will
generate the same error that test program X creates, but more often. To
find out if my test program is better than the original, so that I can
convince the management that I'm right, I ran my test program to find
out how often I can generate the same error. When I ran my test
program 50 times, I generated the error twice. While this may not
seem much better, I think that I can convince the management to use
my test program instead of the original test program. Am I right?

Exercise:

Problem: Japanese Girls’ Names
by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.”
(The trend might have started around my grandmothers’ generation
and its peak might have been around my mother’s generation.) “Ko”
means “child” in Chinese character. Parents would name their
daughters with “ko” attaching to other Chinese characters which have
meanings that they want their daughters to become, such as Sachiko —
a happy child, Yoshiko — a good child, Yasuko — a healthy child, and
sO on.

However, I noticed recently that only two out of nine of my Japanese
girlfriends at this school have names which end with “ko.” More and
more, parents seem to have become creative, modernized, and,
sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s
generation would have names with “ko” at the end, the proportion has
dropped among my peers. I wrote down all my Japanese friends’, ex-
classmates’, co-workers, and acquaintances’ names that I could
remember. Below are the names. (Some are repeats.) Test to see if the
proportion has dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko,
Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi,

Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi,
Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho,
Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako,
Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika,
Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka,
Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako,
Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko,
Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko,
Yuko, Yuko.

Solution:

ez = —2.99

f0.0014

hDecision: Reject null; Conclusion: p < .70
i (0.4529,0.6582)

Exercise:

Problem: Phillip’s Wish by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.
Again Phillip complained to his mother,
Saying half his friends (including their
brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”
We proceeded to call fifty guys

To see whose prediction would fly.
Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.
Then there’s the other thirty-one,
Who said they’d never have this done.
So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

Exercise:

Problem: The Craven by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.
Pondering over last night’s homework
Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping
Suddenly, there came a tapping.

AS someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”
The teacher said, his voice was deep.

“So a tally I’ve begun to keep
Of every class you nap and snore.
The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,
A percentage less than forty-four.”
This I said and nothing more.

“We'll see,” he said and walked away,
And fifty classes from that day
He counted till the month of May
The classes in which I napped and snored.
The number he found was twenty-four.

At a Significance level of 0.05,
Please tell me am I still alive?
Or did my grade just take a dive
Plunging down beneath the floor?
Upon thee I hereby implore.

Solution:

e e0.57
e f0.7156

e hDecision: Do not reject null
e i (0.3415,0.6185)

Exercise:

Problem:

Toastmasters International cites a report by Gallop Poll that 40% of
Americans fear public speaking. A student believes that less than 40%
of students at her school fear public speaking. She randomly surveys
361 schoolmates and finds that 135 report they fear public speaking.
Conduct a hypothesis test to determine if the percent at her school is
less than 40%. (Source: http:/Atoastmasters.org/artisan/detail.asp?

Exercise:

Problem:

68% of online courses taught at community colleges nationwide were
taught by full-time faculty. To test if 68% also represents California’s
percent for full-time faculty teaching the online classes, Long Beach
City College (LBCC), CA, was randomly selected for comparison. In
the same year, 34 of the 44 online courses LBCC offered were taught
by full-time faculty. Conduct a hypothesis test to determine if 68%
represents CA. NOTE: For more accurate results, use more CA
community colleges and this past year's data. (Sources: Growing by
Degrees by Allen and Seaman; Amit Schitai, Director of Instructional
Technology and Distance Learning, LBCC).

Solution:

e e132

e £0.1873

e hDecision: Do not reject null
e i (0.65,0.90)

Exercise:

Problem:

According to an article in Bloomberg Businessweek, New York City's
most recent adult smoking rate is 14%. Suppose that a survey is
conducted to determine this year’s rate. Nine out of 70 randomly
chosen N.Y. City residents reply that they smoke. Conduct a
hypothesis test to determine if the rate is still 14% or if it has
decreased. (Source: http:/,www.businessweek.com/news/2011-09-
15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html)

Exercise:

Problem:

The mean age of De Anza College students in a previous term was 26.6
years old. An instructor thinks the mean age for online students is older
than 26.6. She randomly surveys 56 online students and finds that the
sample mean is 29.4 with a standard deviation of 2.1. Conduct a
hypothesis test. (Source:

Solution:

e e9.98

¢ f0.0000

e hDecision: Reject null
e i (28.8,30.0)

Exercise:

Problem:

Registered nurses earned an average annual salary of $69,110. For that
same year, a survey was conducted of 41 California registered nurses
to determine if the annual salary is higher than $69,110 for California
nurses. The sample average was $71,121 with a sample standard
deviation of $7,489. Conduct a hypothesis test. (Source:
http://www.bls.gov/oes/current/oes291111.htm)

Exercise:

Problem:

La Leche League International reports that the mean age of weaning a
child from breastfeeding is age 4 to 5 worldwide. In America, most
nursing mothers wean their children much earlier. Suppose a random
survey is conducted of 21 U.S. mothers who recently weaned their
children. The mean weaning age was 9 months (3/4 year) with a
standard deviation of 4 months. Conduct a hypothesis test to determine
if the mean weaning age in the U.S. is less than 4 years old. (Source:
http://www. lalecheleague.org/Law/BAFeb01.html)

Solution:

e e-44.7

e f0.0000

e hDecision: Reject null
e i (0.60,0.90) - in years

Try these multiple choice questions.

Exercise:

Problem:

When a new drug is created, the pharmaceutical company must subject
it to testing before receiving the necessary permission from the Food
and Drug Administration (FDA) to market the drug. Suppose the null
hypothesis is “the drug is unsafe.” What is the Type II Error?

e ATo conclude the drug is safe when in, fact, it is unsafe

¢ BTo not conclude the drug is safe when, in fact, it is safe.

¢ CTo conclude the drug is safe when, in fact, it is safe.

¢ DTo not conclude the drug is unsafe when, in fact, it is unsafe

Solution:

B

The next two questions refer to the following information: Over the past
few decades, public health officials have examined the link between weight
concerns and teen girls smoking. Researchers surveyed a group of 273
randomly selected teen girls living in Massachusetts (between 12 and 15
years old). After four years the girls were surveyed again. Sixty-three (63)
said they smoked to stay thin. Is there good evidence that more than thirty
percent of the teen girls smoke to stay thin?

Exercise:

Problem:The alternate hypothesis is

¢ Ap < 0.30
© Bp < 0.30
© Cp> 0.30
¢ Dp> 0.30

Solution:

D

Exercise:

Problem: After conducting the test, your decision and conclusion are

e AReject H,: There is sufficient evidence to conclude that more
than 30% of teen girls smoke to stay thin.

¢ BDo not reject H,: There is not sufficient evidence to conclude
that less than 30% of teen girls smoke to stay thin.

¢ CDo not reject H,: There is not sufficient evidence to conclude
that more than 30% of teen girls smoke to stay thin.

e DReject H,: There is sufficient evidence to conclude that less
than 30% of teen girls smoke to stay thin.

Solution:

C

The next three questions refer to the following information: A statistics
instructor believes that fewer than 20% of Evergreen Valley College (EVC)
students attended the opening night midnight showing of the latest Harry
Potter movie. She surveys 84 of her students and finds that 11 of attended
the midnight showing.

Exercise:

Problem: An appropriate alternative hypothesis is

¢ Ap= 0.20
¢ Bp > 0.20
¢e Cp< 0.20
¢ Dp < 0.20

Solution:

C

Exercise:

Problem: At a 1% level of significance, an appropriate conclusion is:

e AThere is insufficient evidence to conclude that the percent of
EVC students that attended the midnight showing of Harry Potter
is less than 20%.

e BThere is sufficient evidence to conclude that the percent of EVC
students that attended the midnight showing of Harry Potter is
more than 20%.

e CThere is sufficient evidence to conclude that the percent of EVC
students that attended the midnight showing of Harry Potter is
less than 20%.

¢ DThere is insufficient evidence to conclude that the percent of
EVC students that attended the midnight showing of Harry Potter
is at least 20%.

Solution:

A
Exercise:

Problem:

The Type I error is to conclude that the percent of EVC students who
attended is

e Aat least 20%, when in fact, it is less than 20%.
¢ B20%, when in fact, it is 20%.

e Cless than 20%, when in fact, it is at least 20%.

e Dless than 20%, when in fact, it is less than 20%.

Solution:

C

The next two questions refer to the following information:

It is believed that Lake Tahoe Community College (LTCC) Intermediate
Algebra students get less than 7 hours of sleep per night, on average. A
survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24
hours with a standard deviation of 1.93 hours. At a level of significance of
5%, do LTCC Intermediate Algebra students get less than 7 hours of sleep
per night, on average?

Exercise:

Problem:The distribution to be used for this test is _X ~

¢ A N(7.24,+22)

? ./22
¢ B N(7.24,1.93)
e Cit
¢ Dt2
Solution:
D
Exercise:
Problem:

The Type II error is to not reject that the mean number of hours of
sleep LTCC students get per night is at least 7 when, in fact, the mean
number of hours

e Ais more than 7 hours.
e Bis at most 7 hours.

e Cis at least 7 hours.

e Dis less than 7 hours.

Solution:

D

The next three questions refer to the following information: Previously,
an organization reported that teenagers spent 4.5 hours per week, on
average, on the phone. The organization thinks that, currently, the mean is
higher. Fifteen (15) randomly chosen teenagers were asked how many hours
per week they spend on the phone. The sample mean was 4.75 hours with a
sample standard deviation of 2.0. Conduct a hypothesis test.

Exercise:

Problem:The null and alternate hypotheses are:

AH,:x = 4.5, Haix > 4.5
B Ho: > 4.5 Hai < 4.5
C H,: = 4.75 Hyp > 4.75
D How = 4.5 Hou > 4.5

Solution:

D

Exercise:

Problem:

At a significance level of a = 0.05, what is the correct conclusion?

AThere is enough evidence to conclude that the mean number of
hours is more than 4.75

BThere is enough evidence to conclude that the mean number of
hours is more than 4.5

CThere is not enough evidence to conclude that the mean number
of hours is more than 4.5

DThere is not enough evidence to conclude that the mean number
of hours is more than 4.75

Solution:

C

Exercise:

Problem:The Type I error is:

ATo conclude that the current mean hours per week is higher than
4.5, when in fact, it is higher.

BTo conclude that the current mean hours per week is higher than
4.5, when in fact, it is the same.

¢ CTo conclude that the mean hours per week currently is 4.5,
when in fact, it is higher.

¢ DTo conclude that the mean hours per week currently is no higher
than 4.5, when in fact, it is not higher.

Solution:

B

Review

This module provides an overview of Hypothesis Testing of Single Mean
and Single Proportion as a part of Collaborative Statistics collection
(col10522) by Barbara Illowsky and Susan Dean.

Exercise:

Problem:
Rebecca and Matt are 14 year old twins. Matt’s height is 2 standard
deviations below the mean for 14 year old boys’ height. Rebecca’s

height is 0.10 standard deviations above the mean for 14 year old girls’
height. Interpret this.

e AMatt is 2.1 inches shorter than Rebecca

e BRebecca is very tall compared to other 14 year old girls.
e CRebecca is taller than Matt.

e DMatt is shorter than the average 14 year old boy.

Solution:

D
Exercise:

Problem:

Construct a histogram of the IPO data (see Table of Contents, 14.
Appendix, Data Sets). Use 5 intervals.

Solution:

No solution provided. There are several ways in which the histogram
could be constructed.

The next three exercises refer to the following information: Ninety
homeowners were asked the number of estimates they obtained before
having their homes fumigated. X = the number of estimates.

x Rel. Freq. Cumulative Rel. Freq.

1 0.3
2 0.2
4 0.4
rs) 0.1

Complete the cumulative relative frequency column.
Exercise:

Problem:

Calculate the sample mean (a), the sample standard deviation (b) and
the percent of the estimates that fall at or below 4 (c).

Solution:

e a2.8
e b1.48
e c90%

Exercise:
Problem:

Calculate the median, M, the first quartile, Q1, the third quartile, Q3.
Then construct a boxplot of the data.

Solution:

MSS eOle= 1603 4

Exercise:

Problem: The middle 50% of the data are between and

Solution:

1 and 4

The next three questions refer to the following table: Seventy 5th and
6th graders were asked their favorite dinner.

Fried
Pizza Hamburgers Spaghetti shrimp
oth 15 6 9 0
grader
6th 15 7 10 8
grader
Exercise:
Problem:

Find the probability that one randomly chosen child is in the 6th grade
and prefers fried shrimp.

Solution:

D

Exercise:

Problem: Find the probability that a child does not prefer pizza.

e A 2

Solution:

C
Exercise:

Problem:

Find the probability a child is in the 5th grade given that the child
prefers spaghetti.

eA
° Ba
eae
°Da

Solution:

A

Exercise:

Problem: A sample of convenience is a random sample.

e Atrue

e Bfalse

Solution:

B

Exercise:

Problem: A statistic is a number that is a property of the population.

e Atrue
e Bfalse

Solution:

B

Exercise:

Problem: You should always throw out any data that are outliers.

e Atrue
e Bfalse

Solution:

B
Exercise:

Problem:

Lee bakes pies for a small restaurant in Felton, CA. She generally
bakes 20 pies in a day, on the average. Of interest is the num.ber of
pies she bakes each day

e aDefine the Random Variable X.

e bState the distribution for X.
e cFind the probability that Lee bakes more than 25 pies in any
given day.

Solution:

° b P(20)
e c0.1122

Exercise:

Problem:

Six different brands of Italian salad dressing were randomly selected at
a supermarket. The grams of fat per serving are 7, 7, 9, 6, 8, 5. Assume
that the underlying distribution is normal. Calculate a 95% confidence
interval for the population mean grams of fat per serving of Italian
salad dressing sold in supermarkets.

Solution:

CI: (5.52,8.48)
Exercise:

Problem:

Given: uniform, exponential, normal distributions. Match each to a
statement below.

e amean = median ~ mode
e bmean > median > mode
e cmean = median = mode

Solution:

e auniform

e bexponential
¢ cnormal

Introduction

Student Learning Outcomes
By the end of this chapter, the student should be able to:

e Classify hypothesis tests by type.

e Conduct and interpret hypothesis tests for two population means,
population standard deviations known.

e Conduct and interpret hypothesis tests for two population means,
population standard deviations unknown.

¢ Conduct and interpret hypothesis tests for two population proportions.

e Conduct and interpret hypothesis tests for matched or paired samples.

Introduction

Studies often compare two groups. For example, researchers are interested
in the effect aspirin has in preventing heart attacks. Over the last few years,
newspapers and magazines have reported about various aspirin studies
involving two groups. Typically, one group is given aspirin and the other
group is given a placebo. Then, the heart attack rate is studied over several
years.

There are other situations that deal with the comparison of two groups. For
example, studies compare various diet and exercise programs. Politicians
compare the proportion of individuals from different income brackets who
might vote for them. Students are interested in whether SAT or GRE
preparatory courses really help raise their scores.

In the previous chapter, you learned to conduct hypothesis tests on single
means and single proportions. You will expand upon that in this chapter.
You will compare two means or two proportions to each other. The general
procedure is still the same, just expanded.

To compare two means or two proportions, you work with two groups. The
groups are classified either as independent or matched pairs.
Independent groups mean that the two samples taken are independent, that

is, sample values selected from one population are not related in any way to
sample values selected from the other population. Matched pairs consist of
two samples that are dependent. The parameter tested using matched pairs
is the population mean. The parameters tested using independent groups are
either population means or population proportions.

Note:This chapter relies on either a calculator or a computer to calculate
the degrees of freedom, the test statistics, and p-values. TI-83+ and TI-84
instructions are included as well as the test statistic formulas. When using
the TI-83+/TI-84 calculators, we do not need to separate two population
means, independent groups, population variances unknown into large and
small sample sizes. However, most statistical computer software has the
ability to differentiate these tests.

This chapter deals with the following hypothesis tests:
Independent groups (samples are independent)

e Test of two population means.
e Test of two population proportions.

Matched or paired samples (samples are dependent)

e Becomes a test of one population mean.

Matched or Paired Samples
This module provides an overview of Hypothesis Testing: Matched or Paired Samples as a part of
Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean.

. Simple random sampling is used.

. Sample sizes are often small.

Two measurements (samples) are drawn from the same pair of individuals or objects.

. Differences are calculated from the matched or paired samples.

. The differences form the sample that is used for the hypothesis test.

. The matched pairs have differences that either come from a population that is normal or the number of
differences is sufficiently large so the distribution of the sample mean of differences is approximately
normal.

AunRWNER

In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated.
The differences are the data. The population mean for the differences, jg, is then tested using a Student-t test
for a single population mean with n — 1 degrees of freedom where n is the number of differences.
Equation:

The test statistic (t-score) is:

Example:

Matched or paired samples

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly
selected subjects are shown in the table. The "before" value is matched to an "after" value and the differences
are calculated. The differences have a normal distribution.

Subject: A B C D E F G H

Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6

After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0
Exercise:

Problem:

Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level.

Solution:

Corresponding "before" and "after" values form matched pairs. (Calculate "sfter" - "before").

After Data Before Data Difference

6.8 6.6 0.2
2.4 6.5 -4.1
7.4 9 -1.6
8.5 10.3 -1.8
8.1 11.3 -3.2
6.1 8.1 -2

3.4 6.3 -2.9
2 11.6 -9.6

The data for the test are the differences: {0.2, -4.1, -1.6, -1.8, -3.2, -2, -2.9, -9.6}

The sample mean and sample standard deviation of the differences are: Ga = sl} anal og = BOL
Verify these values.

Let jg be the population mean for the differences. We use the subscript d to denote "differences."

Random Variable: X, = the mean difference of the sensory measurements
Equation:

Ay: pa > 9
There is no improvement. (jg is the population mean of the differences.)
Equation:

Ay: pa< 0

There is improvement. The score should be lower after hypnotism so the difference ought to be negative
to indicate improvement.

Distribution for the test: The distribution is a student-t with df = n — 1 = 8 — 1 = 7. Use tz. (Notice
that the test is for a single population mean.)

Calculate the p-value using the Student-t distribution: p-value = 0.0095

Graph:

p-value = 0.0095

—3.13 0
From H,, Hd 2 0

X q is the random variable for the differences.

The sample mean and sample standard deviation of the differences are:

Ai) = =S)o1l3}

$q= 2.91

Compare a and the p-value: a = 0.05 and p-value = 0.0095. a > p-value.

Make a decision: Since a > p-value, reject Ho.

This means that q < O and there is improvement.

Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude

that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective
in reducing pain.

Note:For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time (after
- before) and put the differences into a list or you can put the after data into a first list and the before
data into a second list. Then go to a third list and arrow up to the name. Enter 1st list name - 2nd list
name. The calculator will do the subtraction and you will have the differences in the third list.

Note:TI-83+ and TI-84: Use your list of differences as the data. Press

STAT

and arrow over to
TESTS

. Press
2:T-Test

. Arrow over to
Data

and press

ENTER

. Arrow down and enter

0

for 49, the name of the list where you put the data, and
1

for Freq:. Arrow down to

U

: and arrow over to

<

[o. Press

ENTER

. Arrow down to

Calculate

and press

ENTER

. The p-value is 0.0094 and the test statistic is -3.04. Do these instructions again except arrow to
Draw

(instead of

Calculate

). Press

ENTER

Example:

A college football coach was interested in whether the college's strength development class increased his
players' maximum lift (in pounds) on the bench press exercise. He asked 4 of his players to participate in a
study. The amount of weight they could each lift was recorded before they took the strength development
class. After completing the class, the amount of weight they could each lift was again measured. The data are
as follows:

Weight (in pounds) Player 1 Player 2 Player 3 Player 4
Amount of weighted lifted prior to the class 205 241 338 368

Amount of weight lifted after the class 295 252 330 360

The coach wants to know if the strength development class makes his players stronger, on average.
Exercise:

Problem:

Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to
the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8,
-8}. The differences have a normal distribution.

Using the differences data, calculate the sample mean and the sample standard deviation.

iba = Pilg} Sq = 46.7

Using the difference data, this becomes a test of a single (fill in the blank).

Define the random variable: X , = mean difference in the maximum lift per player.

The distribution for the hypothesis test is ¢3.

Ay: wa <0 A, : pg > 0

Graph:

p-value = 0.2150

21.3

Calculate the p-value: The p-value is 0.2150

Decision: If the level of significance is 5%, the decision is to not reject the null hypothesis because
a < p-value.

What is the conclusion?
Solution:

means; At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude
that the strength development class helped to make the players stronger, on average.

Example:

Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their
dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal
distances with either hand. The following data was collected.

Distance
(in feet) Student Student Student Student Student Student Student
using 1 2 3 4 5 6 7

Dominant

rand 30 26 34 17 19 26 20

Weaker

rena 28 14 27 18 17 26 16

Exercise:

Problem:

Conduct a hypothesis test to determine whether the mean difference in distances between the children's
dominant versus weaker hands is significant.

Note:use a t-test on the difference data. Assume the differences have a normal distribution. The random
variable is the mean difference.

Note:The test statistic is 2.18 and the p-value is 0.0716.

What is your conclusion?

Solution:

Ao: fa equals 0; Hz: 4g does not equal 0; Do not reject the null; At a 5% significance level, from the
sample data, there is not sufficient evidence to conclude that the mean difference in distances between the
children's dominant versus weaker hands is significant (there is not sufficient evidence to show that the
children could push the shot-put further with their dominant hand). Alpha and the p-value are close so the
test is not strong.

Comparing Two Independent Population Means with Unknown Population
Standard Deviations

This module provides an overview of Comparing Two Independent
Population Means with Unknown Population Standard Deviations as a part
of Collaborative Statistics collection (col10522) by Barbara Illowsky and
Susan Dean.

1. The two independent samples are simple random samples from two
distinct populations.

2. Both populations are normally distributed with the population means
and standard deviations unknown unless the sample sizes are greater
than 30. In that case, the populations need not be normally distributed.

Note:The test comparing two independent population means with
unknown and possibly unequal population standard deviations is called the
Aspin-Welch t-test. The degrees of freedom formula was developed by
Aspin-Welch.

The comparison of two population means is very common. A difference
between the two samples depends on both the means and the standard
deviations. Very different means can occur by chance if there is great
variation among the individual samples. In order to account for the
variation, we take the difference of the sample means, X, - X92 , and divide
by the standard error (shown below) in order to standardize the difference.
The result is a t-score test statistic (shown below).

Because we do not know the population standard deviations, we estimate
them using the two sample standard deviations from our independent
samples. For the hypothesis test, we calculate the estimated standard

deviation, or standard error, of the difference in sample means, Diced

Equation:
The standard error is:

The test statistic (t-score) is calculated as follows:

Equation:
t-score
(1 — £2) — (wa — be)
($1)? 4 (S2)?
ny ne
where:

e sj, and Sg, the sample standard deviations, are estimates of 7, and oo,
respectively.

¢ o, and o>» are the unknown population standard deviations.

e¢ x, and £2 are the sample means. jz; and j/2 are the population means.

The degrees of freedom (df) is a somewhat complicated calculation.
However, a computer or calculator calculates it easily. The dfs are not
always a whole number. The test statistic calculated above is approximated
by the student's-t distribution with dfs as follows:
Equation:

Degrees of freedom

When both sample sizes n1 and 2 are five or larger, the student's-t
approximation is very good. Notice that the sample variances s;7 and s2”
are not pooled. (If the question comes up, do not pool the variances.)

Note: It is not necessary to compute this by hand. A calculator or computer
easily computes it.

Example:

Independent groups

The average amount of time boys and girls ages 7 through 11 spend
playing sports each day is believed to be the same. An experiment is done,
data is collected, resulting in the table below. Both populations have a
normal distribution.

Average Number of Sample
Sample Hours Playing Sports Standard
Size Per Day Deviation
Girls 9 2 hours V0.75
Boys 16 3.2 hours 1.00
Exercise:
Problem:

Is there a difference in the mean amount of time boys and girls ages 7
through 11 play sports each day? Test at the 5% level of significance.

Solution:

The population standard deviations are not known. Let g be the
subscript for girls and 6 be the subscript for boys. Then, jz, is the

population mean for girls and py is the population mean for boys.
This is a test of two independent groups, two population means.

Random variable: X, — X;, = difference in the sample mean amount
of time girls and boys play sports each day.

Fy: [bg = be Lg — Lp = 9
Hq: bg F bb Mg — Hy #0
The words "the same" tell you H, has an "=". Since there are no

other words to indicate H,, then assume "is different." This is a two-
tailed test.

Distribution for the test: Use ta where df is calculated using the df
formula for independent groups, two population means. Using a

calculator, df is approximately 18.8462. Do not pool the variances.

Calculate the p-value using a student's-t distribution: p-value =
0.0054

Graph:

l l
> (p-value) = 0.0028 5 (p-value) = 0.0028

f= ie

Seal

SO Lg Oh — ee re

Half the p-value is below -1.2 and half is above 1.2.
Make a decision: Since a > p-value, reject H,.

This means you reject 4, = fp. The means are different.

Conclusion: At the 5% level of significance, the sample data show
there is sufficient evidence to conclude that the mean number of hours
that girls and boys aged 7 through 11 play sports per day is different
(mean number of hours boys aged 7 through 11 play sports per day is
greater than the mean number of hours played by girls OR the mean
number of hours girls aged 7 through 11 play sports per day is greater
than the mean number of hours played by boys).

Note: TI-83+ and TI-84: Press
STAT

. Arrow over to

TLESES

and press

4:2-SampTTest

. Arrow over to Stats and press
ENTER

. Arrow down and enter

2

for the first sample mean,

V0.75

for Sx1,

9

for n1,

Se

for the second sample mean,
1

for Sx2, and

16

for n2. Arrow down to p11: and arrow to
does not equal

2. Press

ENTER

. Arrow down to Pooled: and
No

. Press

ENTER

. Arrow down to
Calculate

and press

ENTER

. The p-value is p = 0.0054, the dfs are approximately 18.8462, and
the test statistic is -3.14. Do the procedure again but instead of
Calculate do Draw.

Example:

A study is done by a community group in two neighboring colleges to
determine which one graduates students with more math classes. College A
samples 11 graduates. Their average is 4 math classes with a standard
deviation of 1.5 math classes. College B samples 9 graduates. Their
average is 3.5 math classes with a standard deviation of 1 math class. The
community group believes that a student who graduates from college A
has taken more math classes, on the average. Both populations have a
normal distribution. Test at a 1% significance level. Answer the following
questions.

Exercise:

Problem:Is this a test of two means or two proportions?
Solution:

two means

Exercise:

Problem:

Are the populations standard deviations known or unknown?

Solution:

unknown

Exercise:

Problem: Which distribution do you use to perform the test?

Solution:

student's-t

Exercise:

Problem: What is the random variable?
Solution:
X4— XB
Exercise:
Problem: What are the null and alternate hypothesis?
Solution:

. Hy: aS bes
° Ha: a> MB

Exercise:

Problem:Is this test right, left, or two tailed?

Solution:

right

Exercise:

Problem: What is the p-value?

Solution:

0.1928

Exercise:

Problem:Do you reject or not reject the null hypothesis?

Solution:

Do not reject.

Conclusion:

At the 1% level of significance, from the sample data, there is not
sufficient evidence to conclude that a student who graduates from college
A has taken more math classes, on the average, than a student who
graduates from college B.

Glossary

Degrees of Freedom (df)
The number of objects in a sample that are free to vary.

Standard Deviation
A number that is equal to the square root of the variance and measures
how far data values are from their mean. Notation: s for sample
standard deviation and o for population standard deviation.

Variable (Random Variable)
A characteristic of interest in a population being studied. Common
notation for variables are upper case Latin letters X, Y, Z,...; common
notation for a specific value from the domain (set of all possible values
of a variable) are lower case Latin letters x, y, z,.... For example, if X
is the number of children in a family, then x represents a specific
integer 0, 1, 2, 3, .... Variables in statistics differ from variables in
intermediate algebra in two following ways.

e The domain of the random variable (RV) is not necessarily a
numerical set; the domain may be expressed in words; for
example, if X = hair color then the domain is {black, blond, gray,
green, orange}.

e We can tell what specific value x of the Random Variable X takes
only after performing the experiment.

Comparing Two Independent Population Proportions

1. The two independent samples are simple random samples that are
independent.

2. The number of successes is at least five and the number of failures is at
least five for each of the samples.

Comparing two proportions, like comparing two means, is common. If two
estimated proportions are different, it may be due to a difference in the
populations or it may be due to chance. A hypothesis test can help
determine if a difference in the estimated proportions (P’, — P'g) reflects a

difference in the population proportions.

The difference of two proportions follows an approximate normal
distribution. Generally, the null hypothesis states that the two proportions
are the same. That is, H, : p4 = pp. To conduct the test, we use a pooled
proportion, pe.
Equation:

The pooled proportion is calculated as follows:

LAtTXLB
NA+nNB

c —

Equation:
The distribution for the differences is:

1 1
Pra — Ptp-N 0, >: (1 =p.) (= ie =|

Equation:
The test statistic (z-score) is:

_ (p!4 — ple) — (pa — ps)

pe-(1-pe)- (2 +2)

Example:

Two population proportions

Two types of medication for hives are being tested to determine if there is a
difference in the proportions of adult patient reactions. Twenty out of a
random sample of 200 adults given medication A still had hives 30
minutes after taking the medication. Twelve out of another random
sample of 200 adults given medication B still had hives 30 minutes after
taking the medication. Test at a 1% level of significance.

Determining the solution

This is a test of 2 population proportions.
Exercise:

Problem: How do you know?

Solution:

The problem asks for a difference in proportions.

Let A and B be the subscripts for medication A and
medication B. Then py, and pz are the desired population
proportions.

Random Variable:

P’, — P’p = difference in the proportions of adult patients
who did not react after 30 minutes to medication A and
medication B.

Jel NW =o pA — pa =U
Hy: pa # PB PA— pp #FO
The words "is a difference" tell you the test is two-tailed.

Distribution for the test: Since this is a test of two
binomial population proportions, the distribution is normal:

_ fates — 204127 ion
Pe = nating = 2001200 — 9-08 1—pe= 0.92

Therefore,
P’4— Pp 0, \/ (0.08) - (0.92) - (so + say)
P’ 4 — P’z follows an approximate normal distribution.

Calculate the p-value using the normal distribution: p-
value = 0.1404.

Estimated proportion for group A:
ee ee all
PA 7a = 200

Estimated proportion for group B:
Se] | =a Se

1
1... a I
5 eee = ee = (p-value) = 0.0702

~0.04 0 0.04 A
From Ho, pa - pgp = 9.

Po Pos — Oe 0.06. — 004

Half the p-value is below -0.04 and half is above 0.04.

Compare a and the p-value: a = 0.01 and the
p-value = 0.1404. a < p-value.

Make a decision: Since a@ < p-value, do not reject H.

Conclusion: At a 1% level of significance, from the sample
data, there is not sufficient evidence to conclude that there is
a difference in the proportions of adult patients who did not

react after 30 minutes to medication A and medication B.

Note: TI-83+ and TI-84: Press
STAT

. Arrow over to

MEous

and press
6:2-PropZTest

. Arrow down and enter
20

for xl,

200

for nl,

aL.

for x2, and

200

for n2. Arrow down to
p1

: and arrow to

not equal p2

. Press

ENTER

. Arrow down to
Calculate

and press

ENTER

. The p-value is p = 0.1404 and the test statistic is 1.47.
Do the procedure again but instead of

Calculate
do

Draw

Summary of Types of Hypothesis Tests
Two Population Means

e Populations are independent and population standard deviations are
unknown.

e Populations are independent and population standard deviations are
known (not likely).

Matched or Paired Samples

¢ Two samples are drawn from the same set of objects.
e Samples are dependent.

Two Population Proportions

e Populations are independent.

Homework

For questions [link] - [link], indicate which of the following choices best identifies
the hypothesis test.

e Alndependent group means, population standard deviations and/or variances
known

e Blndependent group means, population standard deviations and/or variances
unknown

e CMatched or paired samples

e DSingle mean

e E2 proportions

e FSingle proportion

Exercise:
Problem:
A powder diet is tested on 49 people and a liquid diet is tested on 36 different
people. The population standard deviations are 2 pounds and 3 pounds,

respectively. Of interest is whether the liquid diet yields a higher mean weight
loss than the powder diet.

Solution:

A
Exercise:
Problem:
A new chocolate bar is taste-tested on consumers. Of interest is whether the

proportion of children that like the new chocolate bar is greater than the
proportion of adults that like it.

Exercise:
Problem:
The mean number of English courses taken in a two-year time period by male

and female college students is believed to be about the same. An experiment is
conducted and data are collected from 9 males and 16 females.

Solution:

B
Exercise:
Problem:
A football league reported that the mean number of touchdowns per game was

5. A study is done to determine if the mean number of touchdowns has
decreased.

Exercise:
Problem:
A study is done to determine if students in the California state university
system take longer to graduate than students enrolled in private universities.
100 students from both the California state university system and private

universities are surveyed. From years of research, it is known that the
population standard deviations are 1.5811 years and 1 year, respectively.

Solution:

A
Exercise:

Problem:

According to a YWCA Rape Crisis Center newsletter, 75% of rape victims
know their attackers. A study is done to verify this.

Exercise:

Problem:

According to a recent study, U.S. companies have an mean maternity-leave of
six weeks.

Solution:

D

Exercise:

Problem:

A recent drug survey showed an increase in use of drugs and alcohol among
local high school students as compared to the national percent. Suppose that a
survey of 100 local youths and 100 national youths is conducted to see if the
proportion of drug and alcohol use is higher locally than nationally.

Exercise:
Problem:

A new SAT study course is tested on 12 individuals. Pre-course and post-
course scores are recorded. Of interest is the mean increase in SAT scores.

Solution:

C
Exercise:

Problem:

University of Michigan researchers reported in the Journal of the National
Cancer Institute that quitting smoking is especially beneficial for those under
age 49. In this American Cancer Society study, the risk (probability) of dying
of lung cancer was about the same as for those who had never smoked.

Note:For each of the word problems, use a solution sheet to do the hypothesis test.
The solution sheet is found in 14. Appendix (online book version: the link is
"Solution Sheets"; PDF book version: look under 14.5 Solution Sheets). Please feel
free to make copies of the solution sheets. For the online version of the book, it is
suggested that you copy the .doc or the .pdf files.

Note:If you are using a student's-t distribution for a homework problem below,
including for paired data, you may assume that the underlying population is
normally distributed. (In general, you must first prove that assumption, though.)

Exercise:

Problem:

A powder diet is tested on 49 people and a liquid diet is tested on 36 different
people. Of interest is whether the liquid diet yields a higher mean weight loss
than the powder diet. The powder diet group had an mean weight loss of 42
pounds with a standard deviation of 12 pounds. The liquid diet group had an
mean weight loss of 45 pounds with a standard deviation of 14 pounds.

Solution:

e d t6g.44

e e-1.04

e f0.1519

e¢ hDecision: Do not reject null

Exercise:

Problem:

The mean number of English courses taken in a two-year time period by male
and female college students is believed to be about the same. An experiment is
conducted and data are collected from 29 males and 16 females. The males
took an average of 3 English courses with a standard deviation of 0.8. The
females took an average of 4 English courses with a standard deviation of 1.0.
Are the means statistically the same?

Exercise:

Problem:

A study is done to determine if students in the California state university
system take longer to graduate, on average, than students enrolled in private
universities. 100 students from both the California state university system and
private universities are surveyed. Suppose that from years of research, it is
known that the population standard deviations are 1.5811 years and 1 year,
respectively. The following data are collected. The California state university
system students took on average 4.5 years with a standard deviation of 0.8.
The private university students took on average 4.1 years with a standard
deviation of 0.3.

Solution:

Standard Normal

eez—2.14
e £0.0163
e hDecision: Reject null when a = 0.05; Do not reject null when a = 0.01

Exercise:
Problem:
A new SAT study course is tested on 12 individuals. Pre-course and post-

course scores are recorded. Of interest is the mean increase in SAT scores. The
following data are collected:

Pre-course score Post-course score
1200 1300
960 920
1010 1100
840 880
1100 1070
1250 1320
860 860
1330 1370
790 770
990 1040

1110 1200

740 850

Exercise:

Problem:

A recent drug survey showed an increase in use of drugs and alcohol among
local high school seniors as compared to the national percent. Suppose that a
survey of 100 local seniors and 100 national seniors is conducted to see if the
proportion of drug and alcohol use is higher locally than nationally. Locally, 65
seniors reported using drugs or alcohol within the past month, while 60
national seniors reported using them.

Solution:

¢ e0.73
¢ £0.2326
e hDecision: Do not reject null

Exercise:

Problem:

A student at a four-year college claims that mean enrollment at four-year
colleges is higher than at two-year colleges in the United States. Two surveys
are conducted. Of the 35 two-year colleges surveyed, the mean enrollment
was 5068 with a standard deviation of 4777. Of the 35 four-year colleges
surveyed, the mean enrollment was 5466 with a standard deviation of 8191.
(Source: Microsoft Bookshelf)

Exercise:

Problem:

A study was conducted by the U.S. Army to see if applying antiperspirant to
soldiers’ feet for a few days before a major hike would help cut down on the
number of blisters soldiers had on their feet. In the experiment, for three nights
before they went on a 13-mile hike, a group of 328 West Point cadets put an
alcohol-based antiperspirant on their feet. A “control group” of 339 soldiers
put on a similar, but inactive, preparation on their feet. On the day of the hike,
the temperature reached 83° F. At the end of the hike, 21% of the soldiers who
had used the antiperspirant and 48% of the control group had developed foot
blisters. Conduct a hypothesis test to see if the proportion of soldiers using the
antiperspirant was significantly lower than the control group. (Source: U.S.
Army study reported in Journal of the American Academy of Dermatologists)

Solution:

e e-7.33
e f0
e hDecision: Reject null

Exercise:

Problem:

We are interested in whether the proportions of female suicide victims for ages
15 to 24 are the same for the white and the black races in the United States. We
randomly pick one year, 1992, to compare the races. The number of suicides
estimated in the United States in 1992 for white females is 4930. 580 were
aged 15 to 24. The estimate for black females is 330. 40 were aged 15 to 24.
We will let female suicide victims be our population. (Source: the National
Center for Health Statistics, U.S. Dept. of Health and Human Services)

Exercise:

Problem:

At Rachel’s 11th birthday party, 8 girls were timed to see how long (in
seconds) they could hold their breath in a relaxed position. After a two-minute
rest, they timed themselves while jumping. The girls thought that the mean
difference between their jumping and relaxed times would be 0. Test their
hypothesis.

Relaxed time (seconds) Jumping time (seconds)

26 21

47 40

30 28

22 21

23 25

45 43

37 35

29 a2
Solution:

° dt7

e e-1.51

e £0.1755

e hDecision: Do not reject null

Exercise:

Problem:

Elizabeth Mjelde, an art history professor, was interested in whether the value

larger+smaller dimension
larger dimension

Whitney Exhibit for works from 1900 — 1919 as for works from 1920 — 1942.
37 early works were sampled. They averaged 1.74 with a standard deviation of
0.11. 65 of the later works were sampled. They averaged 1.746 with a standard
deviation of 0.1064. Do you think that there is a significant difference in the
Golden Ratio calculation? (Source: data from Whitney Exhibit on loan to San
Jose Museum of Art)

from the Golden Ratio formula, ( ) was the same in the

Exercise:

Problem:

One of the questions in a study of marital satisfaction of dual—career couples
was to rate the statement, “I’m pleased with the way we divide the
responsibilities for childcare.” The ratings went from 1 (strongly agree) to 5
(strongly disagree). Below are ten of the paired responses for husbands and
wives. Conduct a hypothesis test to see if the mean difference in the husband’s
versus the wife’s satisfaction level is negative (meaning that, within the
partnership, the husband is happier than the wife).

Wife’s
score

Husband’s
score

Solution:

e dtg

e et = —1.86

¢ £0.0479

e hDecision: Reject null, but run another test

Exercise:
Problem:
Ten individuals went on a low—fat diet for 12 weeks to lower their cholesterol.

Evaluate the data below. Do you think that their cholesterol levels were
significantly lowered?

Starting cholesterol level Ending cholesterol level

140 140
220 230
110 120
240 220
200 190
180 150
190 200
360 300
280 300
260 240
Exercise:
Problem:

Mean entry level salaries for college graduates with mechanical engineering
degrees and electrical engineering degrees are believed to be approximately
the same. (Source: http:// www. graduatingengineer.com). A recruiting office
thinks that the mean mechanical engineering salary is actually lower than the
mean electrical engineering salary. The recruiting office randomly surveys 50
entry level mechanical engineers and 60 entry level electrical engineers. Their
mean salaries were $46,100 and $46,700, respectively. Their standard
deviations were $3450 and $4210, respectively. Conduct a hypothesis test to
determine if you agree that the mean entry level mechanical engineering salary
is lower than the mean entry level electrical engineering salary.

Solution:

e dtiog

e ef = —0.82
¢ f0.2066
e hDecision: Do not reject null

Exercise:

Problem:

A recent year was randomly picked from 1985 to the present. In that year,
there were 2051 Hispanic students at Cabrillo College out of a total of 12,328
students. At Lake Tahoe College, there were 321 Hispanic students out of a
total of 2441 students. In general, do you think that the percent of Hispanic
students at the two colleges is basically the same or different? (Source:
Chancellor’s Office, California Community Colleges, November 1994)

Exercise:
Problem:
Eight runners were convinced that the mean difference in their individual times
for running one mile versus race walking one mile was at most 2 minutes.

Below are their times. Do you agree that the mean difference is at most 2
minutes?

Running time (minutes) Race walking time (minutes)
5.1 yf!

5.6 9.2

6.2 10.4

4.8 6.9

ipa 8.9

4.2 9.5

6.1 9.4

4.4 79

Solution:

° dt7

e et = 2.9850

e £0.0102

e hDecision: Reject null; There is sufficient evidence to conclude that the
mean difference is more than 2 minutes.

Exercise:

Problem:

Marketing companies have collected data implying that teenage girls use more
ring tones on their cellular phones than teenage boys do. In one particular
study of 40 randomly chosen teenage girls and boys (20 of each) with cellular
phones, the mean number of ring tones for the girls was 3.2 with a standard
deviation of 1.5. The mean for the boys was 1.7 with a standard deviation of
0.8. Conduct a hypothesis test to determine if the means are approximately the
same or if the girls’ mean is higher than the boys’ mean.

Exercise:

Problem:

While her husband spent 2% hours picking out new speakers, a statistician
decided to determine whether the percent of men who enjoy shopping for
electronic equipment is higher than the percent of women who enjoy shopping
for electronic equipment. The population was Saturday afternoon shoppers.
Out of 67 men, 24 said they enjoyed the activity. 8 of the 24 women surveyed
claimed to enjoy the activity. Interpret the results of the survey.

Solution:

e e0.22
¢ £0.4133
e hDecision: Do not reject null

Exercise:

Problem:

We are interested in whether children’s educational computer software costs
less, on average, than children’s entertainment software. 36 educational
software titles were randomly picked from a catalog. The mean cost was
$31.14 with a standard deviation of $4.69. 35 entertainment software titles
were randomly picked from the same catalog. The mean cost was $33.86 with
a standard deviation of $10.87. Decide whether children’s educational software
costs less, on average, than children’s entertainment software. (Source:
Educational Resources, December catalog)

Exercise:

Problem:

Parents of teenage boys often complain that auto insurance costs more, on
average, for teenage boys than for teenage girls. A group of concerned parents
examines a random sample of insurance bills. The mean annual cost for 36
teenage boys was $679. For 23 teenage girls, it was $559. From past years, it is
known that the population standard deviation for each group is $180.
Determine whether or not you believe that the mean cost for auto insurance for
teenage boys is greater than that for teenage girls.

Solution:

°© ez= 2.50
¢ £0.0063
e hDecision: Reject null

Exercise:

Problem:

A group of transfer bound students wondered if they will spend the same mean
amount on texts and supplies each year at their four-year university as they
have at their community college. They conducted a random survey of 54
students at their community college and 66 students at their local four-year
university. The sample means were $947 and $1011, respectively. The
population standard deviations are known to be $254 and $87, respectively.
Conduct a hypothesis test to determine if the means are statistically the same.

Exercise:

Problem:

Joan Nguyen recently claimed that the proportion of college—age males with at
least one pierced ear is as high as the proportion of college—age females. She
conducted a survey in her classes. Out of 107 males, 20 had at least one
pierced ear. Out of 92 females, 47 had at least one pierced ear. Do you believe
that the proportion of males has reached the proportion of females?

Solution:

e e-4.82
e f0
e hDecision: Reject null

Exercise:

Problem:

Some manufacturers claim that non-hybrid sedan cars have a lower mean
miles per gallon (mpg) than hybrid ones. Suppose that consumers test 21
hybrid sedans and get a mean of 31 mpg with a standard deviation of 7 mpg.
Thirty-one non-hybrid sedans get a mean of 22 mpg with a standard deviation
of 4 mpg. Suppose that the population standard deviations are known to be 6
and 3, respectively. Conduct a hypothesis test to the manufacturers claim.

Questions [link] — [link] refer to the Terri Vogel’s data set (see Table of Contents).
Exercise:

Problem:

Using the data from Lap 1 only, conduct a hypothesis test to determine if the
mean time for completing a lap in races is the same as it is in practices.

Solution:
¢ dt20.32
e e-4.70

¢ f0.0001
e hDecision: Reject null

Exercise:

Problem: Repeat the test in [link], but use Lap 5 data this time.
Exercise:

Problem:
Repeat the test in [link], but this time combine the data from Laps 1 and 5.
Solution:

e dt40.94

e e-5.08

e f0

e hDecision: Reject null

Exercise:
Problem:
In 2 — 3 complete sentences, explain in detail how you might use Terri Vogel’s

data to answer the following question. “Does Terri Vogel drive faster in races
than she does in practices?”

Exercise:

Problem:

Is the proportion of race laps Terri completes slower than 130 seconds less
than the proportion of practice laps she completes slower than 135 seconds?

Solution:
e e-0.9223

e £0.1782
e hDecision: Do not reject null

Exercise:

Problem:"To Breakfast or Not to Breakfast?" by Richard Ayore

In the American society, birthdays are one of those days that everyone looks
forward to. People of different ages and peer groups gather to mark the 18th,

20th, ... birthdays. During this time, one looks back to see what he or she had
achieved for the past year, and also focuses ahead for more to come.

If, by any chance, I am invited to one of these parties, my experience is always
different. Instead of dancing around with my friends while the music is
booming, I get carried away by memories of my family back home in Kenya. I
remember the good times I had with my brothers and sister while we did our
daily routine.

Every morning, I remember we went to the shamba (garden) to weed our
crops. I remember one day arguing with my brother as to why he always
remained behind just to join us an hour later. In his defense, he said that he
preferred waiting for breakfast before he came to weed. He said, “This is why I
always work more hours than you guys!”

And so, to prove his wrong or right, we decided to give it a try. One day we
went to work as usual without breakfast, and recorded the time we could work
before getting tired and stopping. On the next day, we all ate breakfast before
going to work. We recorded how long we worked again before getting tired
and stopping. Of interest was our mean increase in work time. Though not
sure, my brother insisted that it is more than two hours. Using the data below,
solve our problem.

Work hours with breakfast Work hours without breakfast
8 6
i 5
9 5
5 4
9 7

10 7

7 fs)
6 6
9 fs)

Try these multiple choice questions.
For questions [link] — [link], use the following information.

A new AIDS prevention drugs was tried on a group of 224 HIV positive patients.
Forty-five (45) patients developed AIDS after four years. In a control group of 224
HIV positive patients, 68 developed AIDS after four years. We want to test whether
the method of treatment reduces the proportion of patients that develop AIDS after
four years or if the proportions of the treated group and the untreated group stay the
same.

Let the subscript ¢= treated patient and ut= untreated patient.
Exercise:

Problem: The appropriate hypotheses are:

© A Ao:p: < put and Ha:pt > Put
¢ BAo:p; < pur and Ay:p; > put
¢ C Hy:p: = pur and Ha:p: F# Put
« D A 52p5 = Put and Hyp; = Put

Solution:
iD
Exercise:
Problem: If the p -value is 0.0062 what is the conclusion (use a = 0.05 )?

e AThe method has no effect.

¢ BThere is sufficient evidence to conclude that the method reduces the
proportion of HIV positive patients that develop AIDS after four years.

e CThere is sufficient evidence to conclude that the method increases the
proportion of HIV positive patients that develop AIDS after four years.

¢ DThere is insufficient evidence to conclude that the method reduces the
proportion of HIV positive patients that develop AIDS after four years.

Solution:

B
Exercise:

Problem:

Lesley E. Tan investigated the relationship between left-handedness and right-
handedness and motor competence in preschool children. Random samples of
A1 left-handers and 41 right-handers were given several tests of motor skills to
determine if there is evidence of a difference between the children based on
this experiment. The experiment produced the means and standard deviations
shown below. Determine the appropriate test and best distribution to use for
that test.

Left-handed Right-handed
Sample size Al 41
Sample mean 97.5 98.1
Sample standard deviation LS 19.2

e ATwo independent means, normal distribution

¢ BTwo independent means, student's-t distribution

e CMatched or paired samples, student's-t distribution
¢ DTwo population proportions, normal distribution

Solution:

B

For questions [link] — [link], use the following information.

An experiment is conducted to show that blood pressure can be consciously reduced
in people trained in a “biofeedback exercise program.” Six (6) subjects were
randomly selected and the blood pressure measurements were recorded before and
after the training. The difference between blood pressures was calculated

(after — before) producing the following results: xg = —10.2 sq = 8.4. Using
the data, test the hypothesis that the blood pressure has decreased after the training,
Exercise:

Problem: The distribution for the test is

e Ats
e Bts
¢ C N(-10.2,8.4)

°D N(=10:2,77)

Solution:

A

Exercise:

Problem: If a = 0.05, the p-value and the conclusion are

e A0.0014; There is sufficient evidence to conclude that the blood pressure
decreased after the training

¢ BO.0014; There is sufficient evidence to conclude that the blood pressure
increased after the training

¢ €0.0155; There is sufficient evidence to conclude that the blood pressure
decreased after the training

¢ DO0.0155; There is sufficient evidence to conclude that the blood pressure
increased after the training

Solution:

Cc

For questions [link |— [link], use the following information.

The Eastern and Western Major League Soccer conferences have a new Reserve
Division that allows new players to develop their skills. Data for a randomly picked
date showed the following annual goals.

Western Eastern

Los Angeles 9 D.C. United 9
FC Dallas 3 Chicago 8
Chivas USA 4 Columbus 7
Real Salt Lake 3 New England 6
Colorado 4 MetroStars 5
San Jose 4 Kansas City 3

Conduct a hypothesis test to determine if the Western Reserve Division teams score,
on average, fewer goals than the Eastern Reserve Division teams. Subscripts: 1
Western Reserve Division (W); 2 Eastern Reserve Division (E)

Exercise:

Problem: The exact distribution for the hypothesis test is:

e AThe normal distribution.

e BThe student's-t distribution.
e CThe uniform distribution.

e DThe exponential distribution.

Solution:

B

Exercise:

Problem: If the level of significance is 0.05, the conclusion is:

e AThere is sufficient evidence to conclude that the W Division teams
score, on average, fewer goals than the E teams.

e BThere is insufficient evidence to conclude that the W Division teams
score, on average, more goals than the E teams.

¢ CThere is insufficient evidence to conclude that the W teams score, on

average, fewer goals than the E teams score.

DUnable to determine.

Solution:

C

Questions [link] — [link] refer to the following.

Neuroinvasive West Nile virus refers to a severe disease that affects a person’s
nervous system . It is spread by the Culex species of mosquito. In the United States
in 2010 there were 629 reported cases of neuroinvasive West Nile virus out of a
total of 1021 reported cases and there were 486 neuroinvasive reported cases out of
a total of 712 cases reported in 2011. Is the 2011 proportion of neuroinvasive West
Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus
cases? Using a 1% level of significance, conduct an appropriate hypothesis test.
(Source: http:// http:/Awww.cdc.gov/ncidod/dvbid/westnile/index.htm )

e “2011” subscript: 2011 group.
e “2010” subscript: 2010 group

Exercise:

Problem: This is:

e Aa test of two proportions
¢ Ba test of two independent means

¢ Ca test of a single mean
¢ Da test of matched pairs.

Solution:

A

Exercise:

Problem: An appropriate null hypothesis is:

¢ A poo11 < P2010
¢ Bp2011 = P2010
¢ C p2011 < 2010
¢ D p2011 > P2010

Solution:

A
Exercise:

Problem:

The p-value is 0.0022. At a 1% level of significance, the appropriate
conclusion is

e AThere is sufficient evidence to conclude that the proportion of people in
the United States in 2011 that got neuroinvasive West Nile disease is less
than the proportion of people in the United States in 2010 that got
neuroinvasive West Nile disease.

e BThere is insufficient evidence to conclude that the proportion of people
in the United States in 2011 that got neuroinvasive West Nile disease is
more than the proportion of people in the United States in 2010 that got
neuroinvasive West Nile disease.

e CThere is insufficient evidence to conclude that the proportion of people
in the United States in 2011 that got neuroinvasive West Nile disease is
less than the proportion of people in the United States in 2010 that got
neuroinvasive West Nile disease.

e¢ DtThere is sufficient evidence to conclude that the proportion of people in
the United States in 2011 that got neuroinvasive West Nile disease is

more than the proportion of people in the United States in 2010 that got
neuroinvasive West Nile disease.

Solution:

D

Questions [link] and [link] refer to the following:

A golf instructor is interested in determining if her new technique for improving
players’ golf scores is effective. She takes four (4) new students. She records their
18-holes scores before learning the technique and then after having taken her class.
She conducts a hypothesis test. The data are as follows.

ial faa ae Player 4
ci score before 83 78 93 87
Mean score after class 80 80 86 86
Exercise:

Problem: This is:

e Aa test of two independent means
e Ba test of two proportions

¢ Ca test of a single proportion

¢ Da test of matched pairs.

Solution:

i)

Exercise:

Problem: The correct decision is:

e AReject H,
¢ BDo not reject H,

Solution:

B

Questions [link] and [link] refer to the following:

Suppose a Statistics instructor believes that there is no significant difference
between the mean class scores of statistics day students on Exam 2 and Statistics
night students on Exam 2. She takes random samples from each of the populations.
The mean and standard deviation for 35 statistics day students were 75.86 and
16.91. The mean and standard deviation for 37 statistics night students were 75.41
and 19.73. The “day” subscript refers to the statistics day students. The “night”
subscript refers to the statistics night students.

Exercise:

Problem: An appropriate alternate hypothesis for the hypothesis test is:

° A Lday > Hnight
° B Uday < Hnight
eC Uday = Hnight
° D Uday - Hnight

Solution:
D
Exercise:

Problem: A concluding statement is:

e AThere is sufficient evidence to conclude that statistics night students
mean on Exam 2 is better than the statistics day students mean on Exam

Z:

¢ BThere is insufficient evidence to conclude that the statistics day students
mean on Exam 2 is better than the statistics night students mean on Exam
ps

e CThere is insufficient evidence to conclude that there is a significant
difference between the means of the statistics day students and night
students on Exam 2.

¢ DThere is sufficient evidence to conclude that there is a significant
difference between the means of the statistics day students and night
students on Exam 2.

Solution:

C

Review
The next three questions refer to the following information:

In a survey at Kirkwood Ski Resort the following information was
recorded:

0-10 11 - 20 21-40 40+
Ski 10 12 30 8
Snowboard 6 17, 12 5

Sport Participation by Age

Suppose that one person from of the above was randomly selected.
Exercise:

Problem:
Find the probability that the person was a skier or was age 11 — 20.

Solution:

17
100

Exercise:

Problem:

Find the probability that the person was a snowboarder given he/she
was age 21 — 40.

Solution:

12
42

Exercise:

Problem: Explain which of the following are true and which are false.

e aSport and Age are independent events.
e bSki and age 11 — 20 are mutually exclusive events.
e c P(Skiand age 21 — 40) < P(Ski | age 21 — 40)
ed
P(Snowboard or age 0 — 10) < P(Snowboard | age 0 — 10)

Solution:

e aFalse
e bFalse
e clrue
e dFalse

Exercise:
Problem:
The average length of time a person with a broken leg wears a cast is
approximately 6 weeks. The standard deviation is about 3 weeks.
Thirty people who had recently healed from broken legs were

interviewed. State the distribution that most accurately reflects total
time to heal for the thirty people.

Solution:

N(180,16.43)

Exercise:

Problem:

The distribution for _X is Uniform. What can we say for certain about
the distribution for X when n = 1?

e AThe distribution for X is still Uniform with the same mean and
standard dev. as the distribution for X.

e BThe distribution for X is Normal with the different mean and a
different standard deviation as the distribution for X.

e CThe distribution for X is Normal with the same mean but a
larger standard deviation than the distribution for X.

e DtThe distribution for X is Normal with the same mean but a
smaller standard deviation than the distribution for X.

Solution:

A
Exercise:

Problem:

The distribution for X is uniform. What can we say for certain about
the distribution for }* X when n = 50?

¢ AThe distribution for 5* Xis still uniform with the same mean
and standard deviation as the distribution for X.

¢ BtThe distribution for 5} X is Normal with the same mean but a
larger standard deviation as the distribution for X.

¢ CThe distribution for 5) X is Normal with a larger mean and a
larger standard deviation than the distribution for X.

¢ DThe distribution for 5) X is Normal with the same mean but a
smaller standard deviation than the distribution for X.

Solution:

C

The next three questions refer to the following information:

A group of students measured the lengths of all the carrots in a five-pound
bag of baby carrots. They calculated the average length of baby carrots to
be 2.0 inches with a standard deviation of 0.25 inches. Suppose we
randomly survey 16 five-pound bags of baby carrots.

Exercise:

Problem:

State the approximate distribution for X, the distribution for the
average lengths of baby carrots in 16 five-pound bags. X ~

Solution:

_:25_
N(2,~=)

Exercise:

Problem:

Explain why we cannot find the probability that one individual
randomly chosen carrot is greater than 2.25 inches.

Exercise:
Problem: Find the probability that xz is between 2 and 2.25 inches.

Solution:

0.5000

The next three questions refer to the following information:

At the beginning of the term, the amount of time a student waits in line at
the campus store is normally distributed with a mean of 5 minutes and a
standard deviation of 2 minutes.

Exercise:

Problem: Find the 90th percentile of waiting time in minutes.

Solution:
7.6

Exercise:

Problem: Find the median waiting time for one student.

Solution:

5
Exercise:

Problem:

Find the probability that the average waiting time for 40 students is at
least 4.5 minutes.

Solution:

0.9431

Introduction
This module provides an introduction to Chi-Square Distribution as a part
of Collaborative Statistics collection (col10522) by Barbara Illowsky and
Susan Dean.

Student Learning Outcomes
By the end of this chapter, the student should be able to:

e Interpret the chi-square probability distribution as the sample size
changes.

e Conduct and interpret chi-square goodness-of-fit hypothesis tests.

e Conduct and interpret chi-square test of independence hypothesis tests.

e¢ Conduct and interpret chi-square homogeneity hypothesis tests.

e Conduct and interpret chi-square single variance hypothesis tests.

Introduction

Have you ever wondered if lottery numbers were evenly distributed or if
some numbers occurred with a greater frequency? How about if the types of
movies people preferred were different across different age groups? What
about if a coffee machine was dispensing approximately the same amount
of coffee each time? You could answer these questions by conducting a
hypothesis test.

You will now study a new distribution, one that is used to determine the
answers to the above examples. This distribution is called the Chi-square
distribution.

In this chapter, you will learn the three major applications of the Chi-square
distribution:

e The goodness-of-fit test, which determines if data fit a particular
distribution, such as with the lottery example

e The test of independence, which determines if events are independent,
such as with the movie example

e The test of a single variance, which tests variability, such as with the
coffee example

Note: Though the Chi-square calculations depend on calculators or
computers for most of the calculations, there is a table available (see the
Table of Contents 15. Tables). TI-83+ and TI-84 calculator instructions are
included in the text.

Optional Collaborative Classroom Activity

Look in the sports section of a newspaper or on the Internet for some sports
data (baseball averages, basketball scores, golf tournament scores, football
odds, swimming times, etc.). Plot a histogram and a boxplot using your
data. See if you can determine a probability distribution that your data fits.
Have a discussion with the class about your choice.

Notation

This module provides an overview of Chi-Square Distribution Notation as a
part of Collaborative Statistics collection (col10522) by Barbara Illowsky
and Susan Dean.

The notation for the chi-square distribution is:

xX ~X
where degrees of freedom depend on how chi-square is being used. (If
you want to practice calculating chi-square probabilities then use

n . The degrees of freedom for the three major uses are each

calculated differently.)

For the x distribution, the population mean is and the population
standard deviation is

The random variable is shown as y but may be any upper case letter.

The random variable for a chi-square distribution with k degrees of freedom
is the sum of & independent, squared standard normal variables.

y 2 Z, Ti

Facts About the Chi-Square Distribution

1. The curve is nonsymmetrical and skewed to the right.
2. There is a different chi-square curve for each df.

df = 2 df = 24

3. The test statistic for any test is always greater than or equal to zero.
4. When df > 90, the chi-square curve approximates the normal. For X
~ X atid the mean, 2 = df = 1000 and the standard deviation,

o = ¥2-1000 = 44.7. Therefore, X ~ N(1000, 44.7),
approximately.
5. The mean, p, is located just to the right of the peak.

_/7

vy

In the next sections, you will learn about four different applications of the
Chi-Square Distribution. These hypothesis tests are almost always right-
tailed tests. In order to understand why the tests are mostly right-tailed, you
will need to look carefully at the actual definition of the test statistic. Think
about the following while you study the next four sections. If the expected
and observed values are "far" apart, then the test statistic will be "large" and
we will reject in the right tail. The only way to obtain a test statistic very
close to zero, would be if the observed and expected values are very, very
close to each other. A left-tailed test could be used to determine if the fit
were "too good." A "too good" fit might occur if data had been manipulated

or invented. Think about the implications of right-tailed versus left-tailed
hypothesis tests as you learn the applications of the Chi-Square
Distribution.

Goodness-of-Fit Test
This module describes how the chi-square distribution is used to conduct goodness-of-fit
test.

In this type of hypothesis test, you determine whether the data "fit" a particular
distribution or not. For example, you may suspect your unknown data fit a binomial
distribution. You use a chi-square test (meaning the distribution for the hypothesis test is
chi-square) to determine if there is a fit or not. The null and the alternate hypotheses
for this test may be written in sentences or may be stated as equations or
inequalities.

The test statistic for a goodness-of-fit test is:
Equation:

(0-5)
KE

where:

e O = observed values (data)
e F/= expected values (from theory)
¢ k =the number of different data cells or categories

The observed values are the data values and the expected values are the values you
would expect to get if the null hypothesis were true. There are n terms of the form
(O—E)’

7:
The degrees of freedom are df = (number of categories - 1).

The goodness-of-fit test is almost always right tailed. If the observed values and the
corresponding expected values are not close to each other, then the test statistic can get
very large and will be way out in the right tail of the chi-square curve.

Note:The expected value for each cell needs to be at least 5 in order to use this test.

Example:

Absenteeism of college students from math classes is a major concern to math
instructors because missing class appears to increase the drop rate. Suppose that a study
was done to determine if the actual student absenteeism follows faculty perception. The

faculty expected that a group of 100 students would miss class according to the
following chart.

Number absences per term Expected number of students
0-2 50

3-5 30

6-8 f2

9-11 6

Dag 2

A random survey across all mathematics courses was then done to determine the actual
number (observed) of absences in a course. The next chart displays the result of that
survey.

Number absences per term Actual number of students
0-2 35

3-5 40

6-8 20

9-11 1

1 2er 4

Determine the null and alternate hypotheses needed to conduct a goodness-of-fit test.
H,: Student absenteeism fits faculty perception.

The alternate hypothesis is the opposite of the null hypothesis.
H,: Student absenteeism does not fit faculty perception.

Exercise:

Problem:

Can you use the information as it appears in the charts to conduct the goodness-of-

fit test?

Solution:

No. Notice that the expected number of absences for the "12+" entry is less than 5
(it is 2). Combine that group with the "9 - 11" group to create new tables where the
number of students for each entry are at least 5. The new tables are below.

Number absences per term

Number absences per term

Expected number of students
50
30

12

Actual number of students
35
40

20

Exercise:

Problem: What are the degrees of freedom (df)?
Solution:

There are 4 "cells" or categories in each of the new tables.

Gic— sin bet Oivce llc 10 —.4-- ale 95

Example:

Employers particularly want to know which days of the week employees are absent in a
five day work week. Most employers would like to believe that employees are absent
equally during the week. Suppose a random sample of 60 managers were asked on
which day of the week did they have the highest number of employee absences. The
results were distributed as follows:

Monday Tuesday Wednesday Thursday Friday
Number

of 15 12 9 9 15
Absences

Day of the Week Employees were most Absent

Exercise:
Problem:
For the population of employees, do the days for the highest number of absences
occur with equal frequencies during a five day work week? Test at a 5%
significance level.

Solution:

The null and alternate hypotheses are:

¢ H,: The absent days occur with equal frequencies, that is, they fit a uniform
distribution.

e H,: The absent days occur with unequal frequencies, that is, they do not fit a
uniform distribution.

If the absent days occur with equal frequencies, then, out of 60 absent days (the
total in the sample: 15 + 12 +9 +9 + 15 = 60), there would be 12 absences on
Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday.
These numbers are the expected (/) values. The values in the table are the
observed (QO) values or data.

This time, calculate the x? test statistic by hand. Make a chart with the following
headings and fill in the columns:

e Expected (£) values (12, 12, 12, 12, 12)
e Observed (Q) values (15, 12, 9, 9, 15)
(C=)
° (O-E)’
(O-E)’
The last column (-—;-— ) should have 0.75, 0, 0.75, 0.75, 0.75.

Now add (sum) the last column. Verify that the sum is 3. This is the x test
statistic.

To find the p-value, calculate P (x? = OP This test is right-tailed.

(Use a computer or calculator to find the p-value. You should get
p-value = 0.5578.)

The disare the number of cells. — 1 — 5. I — 4)

TI-83+ and TI-84: Press 2nd DISTR. Arrow down to x2cdf. Press ENTER.
Enter (3, 10499, 4). Rounded to 4 decimal places, you should see 0.5578 which
is the p-value.

Next, complete a graph like the one below with the proper labeling and shading.
(You should shade the right tail.)

= ao

x

The decision is to not reject the null hypothesis.
Conclusion: At a 5% level of significance, from the sample data, there is not

sufficient evidence to conclude that the absent days do not occur with equal
frequencies.

Note:TI-83+ and some TI-84 calculators do not have a special program for the test
Statistic for the goodness-of-fit test. The next example (Example 11-3) has the
calculator instructions. The newer TI-84 calculators have in

STAT TESTS
the test

Chi2 GOF

. To run the test, put the observed values (the data) into a first list and the expected
values (the values you expect if the null hypothesis is true) into a second list. Press

STAT
TESTS

and

Chi2 GOF

. Enter the list names for the Observed list and the Expected list. Enter the degrees
of freedom and press

calculate
or
draw

. Make sure you clear any lists before you start. See below.

Note:To Clear Lists in the calculators: Go into

STAT EDIT

and arrow up to the list name area of the particular list. Press

CLEAR

and then arrow down. The list will be cleared. Or, you can press

STAT
and press 4 (for

ClrList

). Enter the list name and press

ENTER

Example:

One study indicates that the number of televisions that American families have is
distributed (this is the given distribution for the American population) as follows:

Number of Televisions

0

over 3

Percent

10

16

55

11

The table contains expected (£) percents.
A random sample of 600 families in the far western United States resulted in the
following data:

Number of Televisions Frequency
0 66
1 119
2 340
3 60
over 3 bs
Total = 600

The table contains observed (O) frequency values.
Exercise:

Problem:
At the 1% significance level, does it appear that the distribution "number of

televisions" of far western United States families is different from the distribution
for the American population as a whole?

Solution:
This problem asks you to test whether the far western United States families
distribution fits the distribution of the American families. This test is always right-

tailed.

The first table contains expected percentages. To get expected (£) frequencies,
multiply the percentage by 600. The expected frequencies are:

Number of Televisions Percent Expected Frequency

0 10 (0.10) - (600) = 60
1 16 (0.16) - (600) = 96
2 55 (0.55) - (600) = 330
3 6 (0.11) - (600) = 66
over 3 8 (0.08) - (600) = 48

Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI
calculators, you can let the calculator do the math. For example, instead of 60, enter
.10*600.

H,: The "number of televisions" distribution of far western United States families
is the same as the "number of televisions" distribution of the American population.

H,: The "number of televisions" distribution of far western United States families
is different from the "number of televisions" distribution of the American

population.

Distribution for the test: x? where df = (the number of cells) —1 =5—1= 4.

Note:df 4 600 — 1

Calculate the test statistic: x? = 29.65

Graph:

p-value = 0.000006 (almost 0)

Probability statement: p-value = P(x” > 29.65) = 0.000006.

Compare a and the p-value:

sa — 0 01
e p-value = 0.000006

So, a > p-value.
Make a decision: Since a > p-value, reject Ho.

This means you reject the belief that the distribution for the far western states is the
same as that of the American population as a whole.

Conclusion: At the 1% significance level, from the data, there is sufficient
evidence to conclude that the "number of televisions" distribution for the far
western United States is different from the "number of televisions" distribution for
the American population as a whole.

Note: TI-83+ and some TI-84 calculators: Press
STAT

and

ENTER

. Make sure to clear lists

Li

L2

, and

L3

if they have data in them (see the note at the end of Example 11-2). Into
L1

, put the observed frequencies

66

, put the expected frequencies

.10*600, .16*600

’

.55*600

’

.11*600

.08*600

. Arrow over to list

L3

and up to the name area
mp gn

. Enter
(L1-L2)42/L2

and

ENTER

. Press

2nd QUIT

. Press

2nd LIST

and arrow over to
MATH

. Press

5

. You should see
"sum" (Enter L3)
. Rounded to 2 decimal places, you should see
29.65

. Press

2nd DISTR

. Press

7
or Arrow down to

fi X2COT.

and press

ENTER

. Enter

(29.65,1E99, 4)

. Rounded to 4 places, you should see

5.77E-6 = .000006

(rounded to 6 decimal places) which is the p-value.
The newer TI-84 calculators have in

STAT TESTS

the test

Chi2 GOF

. To run the test, put the observed values (the data) into a first list and the expected
values (the values you expect if the null hypothesis is true) into a second list. Press

STAT
TESTS

and

Chi2 GOF

. Enter the list names for the Observed list and the Expected list. Enter the degrees
of freedom and press

calculate

or

draw

. Make sure you clear any lists before you start.

Example:
Exercise:

Problem:

Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and
23 TT. Are the coins fair? Test at a 5% significance level.

Solution:

This problem can be set up as a goodness-of-fit problem. The sample space for
flipping two fair coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25
HH, 25 HT, 25 TH, and 25 TT. This is the expected distribution. The question, "Are
the coins fair?" is the same as saying, "Does the distribution of the coins (20 HH,
27 HT, 30 TH, 23 TT) fit the expected distribution?"

Random Variable: Let X = the number of heads in one flip of the two coins. X
takes on the value 0, 1, 2. (There are 0, 1, or 2 heads in the flip of 2 coins.)
Therefore, the number of cells is 3. Since X = the number of heads, the observed
frequencies are 20 (for 2 heads), 57 (for 1 head), and 23 (for 0 heads or both tails).
The expected frequencies are 25 (for 2 heads), 50 (for 1 head), and 25 (for 0 heads
or both tails). This test is right-tailed.

H,: The coins are fair.

H,: The coins are not fair.

Distribution for the test: y3 where df = 3 — 1 = 2.
Calculate the test statistic: x? = 2.14

Graph:

p-value = 0.3430

Probability statement: p-value = P (x? = 2.14) = 0.3430

Compare a and the p-value:

°° a=0.05
¢ p-value = 0.3430

So, a < p-value.
Make a decision: Since a < p-value, do not reject Ho.

Conclusion: There is insufficient evidence to conclude that the coins are not fair.

Note: TI-83+ and some TI- 84 calculators: Press
STAT

and

ENTER

. Make sure you clear lists

Li

if they have data in them. Into
L1
, put the observed frequencies

20

, put the expected frequencies

25

. Arrow over to list

L3

and up to the name area
mpg

. Enter
(L1-L2)A2/L2

and

ENTER

. Press

2nd QUIT

. Press

2nd LIST

and arrow over to
MATH

. Press

5

. You should see

"sum"

Enter L3

. Rounded to 2 decimal places, you should see
2.14

. Press

2nd DISTR

. Arrow down to
7:x2cdf

(or press

7

). Press

ENTER

. Enter

2.14,1E99,2)

. Rounded to 4 places, you should see
. 3430

which is the p-value.

The newer TI-84 calculators have in
STAT TESTS

the test

Chi2 GOF

. To run the test, put the observed values (the data) into a first list and the expected
values (the values you expect if the null hypothesis is true) into a second list. Press

STAT
TESTS

and

Chi2 GOF

. Enter the list names for the Observed list and the Expected list. Enter the degrees
of freedom and press

calculate
or
draw

. Make sure you clear any lists before you start.

Test of Independence
This module describes how the chi-square distribution can be used to test for independence.

Tests of independence involve using a contingency table of observed (data) values. You first
saw a contingency table when you studied probability in the Probability Topics chapter.

The test statistic for a test of independence is similar to that of a goodness-of-fit test:
Equation:

(O— E)?
(ij) £E

where:

¢ O= observed values

e F = expected values

e 2 =the number of rows in the table

e 7 =the number of columns in the table

There are z - 7 terms of the form oa,

A test of independence determines whether two factors are independent or not. You first
encountered the term independence in Chapter 3. As a review, consider the following
example.

Note:The expected value for each cell needs to be at least 5 in order to use this test.

Example:

Suppose A = a speeding violation in the last year and B = a cell phone user while driving. If
A and B are independent then P(A AND B) = P(A)P(B). A AND Bis the event that a
driver received a speeding violation last year and is also a cell phone user while driving.
Suppose, in a study of drivers who received speeding violations in the last year and who uses
cell phones while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding
violation and 685 did not; 305 were cell phone users while driving and 450 were not.

Let y = expected number of drivers that use a cell phone while driving and received speeding
violations.

If A and B are independent, then P(A AND B) = P(A)P(B). By substitution,
y _ 70 , 305

755 755 | 75D

Solve for y: y= BS = 28.3

About 28 people from the sample are expected to be cell phone users while driving and to
receive speeding violations.

In a test of independence, we state the null and alternate hypotheses in words. Since the
contingency table consists of two factors, the null hypothesis states that the factors are
independent and the alternate hypothesis states that they are not independent (dependent).
If we do a test of independence using the example above, then the null hypothesis is:

H,: Being a cell phone user while driving and receiving a speeding violation are independent
events.

If the null hypothesis were true, we would expect about 28 people to be cell phone users
while driving and to receive a speeding violation.

The test of independence is always right-tailed because of the calculation of the test
Statistic. If the expected and observed values are not close together, then the test statistic is
very large and way out in the right tail of the chi-square curve, like goodness-of-fit.

The degrees of freedom for the test of independence are:

df = (number of columns - 1)(number of rows - 1)

The following formula calculates the expected number (£):
ee (row total)(column total)
~ total number surveyed

Example:

In a volunteer group, adults 21 and older volunteer from one to nine hours each week to
spend time with a disabled senior citizen. The program recruits among community college
students, four-year college students, and nonstudents. The following table is a sample of the
adult volunteers and the number of hours they volunteer per week.

1-3 4-6 7-9 Row
Type of Volunteer Hours Hours Hours Total
Community College 1 96 48 955
Students
Four-Year College 96 133 61 290
Students
Nonstudents 91 150 53 294
Column Total 298 379 162 839

Number of Hours Worked Per Week by Volunteer Type (Observed)The table contains
observed (O) values (data).

Exercise:

Problem: Are the number of hours volunteered independent of the type of volunteer?
Solution:

The observed table and the question at the end of the problem, "Are the number of
hours volunteered independent of the type of volunteer?" tell you this is a test of
independence. The two factors are number of hours volunteered and type of
volunteer. This test is always right-tailed.

H,: The number of hours volunteered is independent of the type of volunteer.
H,: The number of hours volunteered is dependent on the type of volunteer.

The expected table is:

Type of Volunteer 1-3 Hours 4-6 Hours 7-9 Hours
Community College Students 90.57 1509 49.24
Four-Year College Students 103.00 131.00 56.00
Nonstudents 104.42 132.81 56.77

Number of Hours Worked Per Week by Volunteer Type (Expected)The table contains
expected (/) values (data).

For example, the calculation for the expected frequency for the top left cell is

__ (row total)(column total) 255.298 __
i eimnbecimesede eae

Calculate the test statistic: x” = 12.99 (calculator or computer)
Distribution for the test: x3
df = (3 columns — 1)(3 rows — 1) = (2)(2) =4

Graph:

p-value = 0.0113

J+

0 12.99

Probability statement: p-value = P(x” > 12.99) = 0.0113

Compare a and the p-value: Since no a is given, assume a = 0.05.
p-value = 0.0113. a > p-value.

Make a decision: Since a > p-value, reject H,. This means that the factors are not
independent.

Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to
conclude that the number of hours volunteered and the type of volunteer are dependent
on one another.

For the above example, if there had been another type of volunteer, teenagers, what
would the degrees of freedom be?

Note:Calculator instructions follow.

TI-83+ and TI-84 calculator: Press the MATRX key and arrow over to EDIT. Press 1:
[A]. Press 3 ENTER 3 ENTER. Enter the table values by row from Example 11-6.
Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS.
Arrow down to C: x2-TEST. Press ENTER. You should see ObServed:[A] and
Expected: [B]. Arrow down to Calculate. Press ENTER. The test statistic is
12.9909 and the p-value = 0.0113. Do the procedure a second time but arrow down to
Draw instead of calculate.

Example:

De Anza College is interested in the relationship between anxiety level and the need to
succeed in school. A random sample of 400 students took a test that measured anxiety level
and need to succeed in school. The table shows the results. De Anza College wants to know
if anxiety level and need to succeed in school are independent events.

Need to

Succeed Med- Med-

in High high Medium low Low Row
School Anxiety Anxiety Anxiety Anxiety Anxiety Total
“gH 35 42 53 15 10 155
Need

ee 48 63 33 a 193
Need

Low

ikea 4 5 11 15 17 52
CIP 95 127 63 58 400
Total

Need to Succeed in School vs. Anxiety Level

Exercise:

Problem:

How many high anxiety level students are expected to have a high need to succeed in
school?

Solution:

The column total for a high anxiety level is 57. The row total for high need to succeed
in school is 155. The sample size or total surveyed is 400.

__ (row total)(column total) 155-57 __
B= “totalsurveyed §§ 400 22.09

The expected number of students who have a high anxiety level and a high need to
succeed in school is about 22.

Exercise:
Problem:

If the two variables are independent, how many students do you expect to have a low
need to succeed in school and a med-low level of anxiety?

Solution:

The column total for a med-low anxiety level is 63. The row total for a low need to
succeed in school is 52. The sample size or total surveyed is 400.

Exercise:

Problem:

__ (row total)(column total) _
a ae total surveyed aa

e b The expected number of students who have a med-low anxiety level and a
low need to succeed in school is about:

Solution:

__ (row total)(column total) —__
Ce total surveyed = 8.19
b8

Glossary

Contingency Table
The method of displaying a frequency distribution as a table with rows and columns to
show how two variables may be dependent (contingent) upon each other. The table
provides an easy way to calculate conditional probabilities.

Summary of Formulas

This module provides a summary on formulas used in Chi-Square
Distribution as a part of Collaborative Statistics collection (col10522) by
Barbara Illowsky and Susan Dean.

The Chi-Square Probability Distribution
jojdi and: ¢ = /2-~- di

Goodness-of-Fit Hypothesis Test

e Use goodness-of-fit to test whether a data set fits a particular
probability distribution.

e The degrees of freedom are number of cells or categories - 1.

oC» , where O = observed values (data), E =

e The test statistic is 3

expected values (from theory), and k = the number of different data
cells or categories.
e The test is right-tailed.

Test of Independence

e Use the test of independence to test whether two factors are
independent or not.

e The degrees of freedom are equal to
(number of columns - 1)(number of rows - 1).
_ 2

e The test statistic is a so-ey where O = observed values, EF’ =

uJ

expected values, 2 = the number of rows in the table, and 7 = the
number of columns in the table.

e The test is right-tailed.

e If the null hypothesis is true, the expected number
ne (row total)(column total)
~ ——— totalsurveyed

Test of Homogeneity

e Use the test for homogeneity to decide if two populations with
unknown distributions have the same distribution as each other.
e The degrees of freedom are equal to number of columns - 1.

cosa —E)
e The test statistic is »’ so-ey where O = observed values, F =
ud
expected values, 2 = the number of rows in the table, and 7 = the
number of columns in the table.
e The test is right-tailed.

e If the null hypothesis is true, the expected number
Be (row total) (column total)
~— — totalsurveyed

Note:The expected value for each cell needs to be at least 5 in order to use
the Goodness-of-Fit, Independence and Homogeneity tests.

Test of a Single Variance

e Use the test to determine variation.
e The degrees of freedom are the number of samples - 1.

9
e The test statistic is a , where n = the total number of data, s? =

sample variance, and o” = population variance.
e The test may be left, right, or two-tailed.

Practice 1: Goodness-of-Fit Test

This module provides a practice on Chi-Square Distribution as a part of
Collaborative Statistics collection (col10522) by Barbara Illowsky and
Susan Dean.

Student Learning Outcomes

e The student will conduct a goodness-of-fit test.

Given

The following data are real. The cumulative number of AIDS cases reported
for Santa Clara County is broken down by ethnicity as follows: (Source:
HIV/AIDS Epidemiology Santa Clara County, Santa Clara County Public
Health Department, May 2011)

Ethnicity Number of Cases
White 2229
Hispanic 1157
Black/African-American 457
Asian, Pacific Islander 232

Total = 4075

The percentage of each ethnic group in Santa Clara County is as follows:

Ethnicity
White
Hispanic

Black/African-
American

Asian, Pacific
Islander

Expected Results

If the ethnicity of AIDS victims followed the ethnicity of the total county

Percentage of total
county population

42.9%

26.7%

2.6%

27.8%

Total = 100%

Number expected
(round to 2 decimal
places)

1748.18

population, fill in the expected number of cases per ethnic group.

Goodness-of-Fit Test

Perform a goodness-of-fit test to determine whether the make-up of AIDS
cases follows the ethnicity of the general population of Santa Clara County.

Exercise:

Problem: H,:

Exercise:

Problem: H,:

Exercise:

Problem: Is this a right-tailed, left-tailed, or two-tailed test?

Exercise:

Problem: degrees of freedom =

Solution:

degrees of freedom = 3

Exercise:

Problem: test statistic =

Solution:

2016.14

Exercise:

Problem: p-value =

Solution:

Rounded to 4 decimal places, the p-value is 0.0000.
Exercise:
Problem:
Graph the situation. Label and scale the horizontal axis. Mark the

mean and test statistic. Shade in the region corresponding to the p-
value.

Let a
Decision:
Reason for the Decision:

Conclusion (write out in complete sentences):

Discussion Question

Exercise:
Problem:
Does it appear that the pattern of AIDS cases in Santa Clara County

corresponds to the distribution of ethnic groups in this county? Why or
why not?

Practice 2: Contingency Tables
This module provides a practice on Chi-Square Distribution as a part of Collaborative Statistics
collection (col10522) by Barbara Illowsky and Susan Dean.

Student Learning Outcomes
¢ The student will conduct a test for independence using contingency tables.

Conduct a hypothesis test to determine if smoking level and ethnicity are independent.

Collect the Data

Copy the data provided in Probability Topics Practice 1: Contingency Tables into the table below.

Smoking

Level African Native Japanese

Per Day American Hawaiian Latino Americans White TOTALS
1-10

11-20

21-30

31+

TOTALS

Smoking Levels by Ethnicity (Observed)

Hypothesis
State the hypotheses.

©. 1137
°e Ay:

Expected Values

Enter expected values in the above below. Round to two decimal places.

Analyze the Data

Calculate the following values:

Exercise:

Problem: Degrees of freedom =

Solution:
12

Exercise:

Problem: test statistic =

Solution:
10301.8
Exercise:
Problem: p-value =

Solution:
0

Exercise:

Problem: Is this a right-tailed, left-tailed, or two-tailed test? Explain why.

Solution:

right

Graph the Data

Exercise:

Problem:

Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in
the region corresponding to the p-value.

Conclusions

State the decision and conclusion (in a complete sentence) for the following preconceived levels of a .
Exercise:

Problem: a

e a Decision:
e b Reason for the decision:
e c Conclusion (write out in a complete sentence):

Solution:

e aReject the null hypothesis
Exercise:

Problem: a

e a Decision:
e b Reason for the decision:
e c Conclusion (write out in a complete sentence):

Homework

This module provides homework on Chi-Square Distribution as a part of Collaborative
Statistics collection (col10522) by Barbara Illowsky and Susan Dean.

Exercise:

Problem:

e aExplain why the “goodness of fit” test and the “test for independence” are
generally right tailed tests.
e blIf you did a left-tailed test, what would you be testing?

Word Problems

For each word problem, use a solution sheet to solve the hypothesis test problem. Go to The
Table of Contents 14. Appendix for the chi-square solution sheet. Round expected frequency
to two decimal places.

Exercise:

Problem:
A 6-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a

hypothesis test to determine if the die is fair. The data below are the result of the 120
rolls.

Face Value Frequency Expected Frequency
1 15
2 29
3 16
4 15
5 30
6 15

Exercise:

Problem:

The marital status distribution of the U.S. male population, age 15 and older, is as shown
below. (Source: U.S. Census Bureau, Current Population Reports)

Marital Status
never married
married
widowed

divorced/separated

Percent Expected Frequency
31.3

56.1

2.5

10.1

Suppose that a random sample of 400 U.S. young adult males, 18 — 24 years old, yielded
the following frequency distribution. We are interested in whether this age group of males
fits the distribution of the U.S. adult population. Calculate the frequency one would
expect when surveying 400 people. Fill in the above table, rounding to two decimal

places.

Marital Status
never married
married
widowed

divorced/separated

Solution:

e aThe data fits the distribution

Frequency

140

¢ bThe data does not fit the distribution

° 63

e e19.27

¢ f0.0002

e hDecision: Reject Null; Conclusion: Data does not fit the distribution.

The next two questions refer to the following information. The columns in the chart below
contain the Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for the
Advanced Placement Examinee Population for that class and the Overall Student Population.
(Source: http://www.collegeboard.com). Suppose the right column contains the result of a
survey of 1000 local students from that year who took an AP Exam.

AP Examinee Overall Student Survey

Race/Ethnicity Population Population Frequency

Asian, Asian American or 10.2% 5.A% 113

Pacific Islander

Black or African 8.2% 14.5% 9A

American

Hispanic or Latino 15.5% 15.9% 136

American Indian or . 2

Alaska Native te ee )

White 59.4% 61.6% 604

Not reported/other 6.1% 1.4% 43
Exercise:

Problem:

Perform a goodness-of-fit test to determine whether the local results follow the
distribution of the U. S. Overall Student Population based on ethnicity.

Exercise:

Problem:

Perform a goodness-of-fit test to determine whether the local results follow the
distribution of U. S. AP Examinee Population, based on ethnicity.

Solution:

e 65
e e13.4
e £0.0199

e gDecision: Reject null when a = 0.05; Conclusion: Local data do not fit the AP
Examinee Distribution. Decision: Do not reject null when a = 0.01; Conclusion:
There is insufficient evidence to conclude that Local data do not fit the AP
Examinee Distribution.

Exercise:

Problem:

The City of South Lake Tahoe, CA, has an Asian population of 1419 people, out of a total
population of 23,609 (Source: U.S. Census Bureau). Suppose that a survey of 1419 self-
reported Asians in Manhattan, NY, area yielded the data in the table below. Conduct a
goodness of fit test to determine if the self-reported sub-groups of Asians in the
Manhattan area fit that of the Lake Tahoe area.

Race

Asian Indian
Chinese
Filipino
Japanese
Korean
Vietnamese

Other

Lake Tahoe Frequency
131

118

1045

80

12

9

24

Manhattan Frequency
174

bor

518

54

29

Zt

66

The next two questions refer to the following information: UCLA conducted a survey of
more than 263,000 college freshmen from 385 colleges in fall 2005. The results of student
expected majors by gender were reported in The Chronicle of Higher Education (2/2/2006).
Suppose a survey of 5000 graduating females and 5000 graduating males was done as a
follow-up last year to determine what their actual major was. The results are shown in the
tables for Exercises 7 and 8. The second column in each table does not add to 100% because
of rounding.

Exercise:

Problem:

Conduct a hypothesis test to determine if the actual college major of graduating females
fits the distribution of their expected majors.

Major Women - Expected Major Women - Actual Major
Arts & Humanities 14.0% 670
Biological Sciences 8.4% 410
Business 13.1% 685
Education 13.0% 650
Engineering 2.6% 145
Physical Sciences 2.6% 125
Professional 18.9% 975
Social Sciences 13.0% 605
Technical 0.4% 15
Other 5.8% 300
Undecided 8.0% 420
Solution:

e cl10

e e11.48

e f0.3214

¢ hDecision: Do not reject null when a = 0.05 and a = 0.01; Conclusion: There is
insufficient evidence to conclude that the distribution of majors by graduating
females does not fit the distribution of expected majors.

Exercise:
Problem:

Conduct a hypothesis test to determine if the actual college major of graduating males fits
the distribution of their expected majors.

Major Men - Expected Major Men - Actual Major
Arts & Humanities 11.0% 600
Biological Sciences 6.7% 330
Business 22.7% 1130
Education 5.8% 305
Engineering 15.6% 800
Physical Sciences 3.6% 175
Professional 9.3% 460
Social Sciences 7.6% 370
Technical 1.8% 90
Other 8.2% 400
Undecided 6.6% 340

Exercise:

Problem:
A recent debate about where in the United States skiers believe the skiing is best

prompted the following survey. Test to see if the best ski area is independent of the level
of the skier.

U.S. Ski Area Beginner Intermediate Advanced
Tahoe 20 30 40
Utah 10 30 60

Colorado 10 40 50

Solution:

e c4

e e10.53

¢ f0.0324

e hDecision: Reject null; Conclusion: Best ski area and level of skier are not
independent.

Exercise:
Problem:
Car manufacturers are interested in whether there is a relationship between the size of car
an individual drives and the number of people in the driver’s family (that is, whether car

size and family size are independent). To test this, suppose that 800 car owners were
randomly surveyed with the following results. Conduct a test for independence.

Family Size Sub & Compact Mid-size Full-size Van & Truck

ul 20 35 40 35

Family Size
2
3-4

5+

Exercise:

Problem:

Sub & Compact
20
20

20

Mid-size Full-size
50 70

50 100

30 70

Van & Truck

80

90

70

College students may be interested in whether or not their majors have any effect on
starting salaries after graduation. Suppose that 300 recent graduates were surveyed as to
their majors in college and their starting salaries after graduation. Below are the data.
Conduct a test for independence.

Major
English
Engineering
Nursing
Business

Psychology

Solution:

c8
e33.55
f0

events.

Exercise:

< $50,000
5

10

10

10

20

$50,000 - $68,999
20
30
15
20

30

$69,000 +
5

60

15

30

20

hDecision: Reject null; Conclusion: Major and starting salary are not independent

Problem:

Some travel agents claim that honeymoon hot spots vary according to age of the bride
and groom. Suppose that 280 East Coast recent brides were interviewed as to where they
spent their honeymoons. The information is given below. Conduct a test for
independence.

Location 20 - 29 30 - 39 40 - 49 50 and over
Niagara Falls 15 25 25 20
Poconos 15 25 25 10
Europe 10 25 15 fs)
Virgin Islands 20 25 15 fs)
Exercise:
Problem:

A manager of a sports club keeps information concerning the main sport in which
members participate and their ages. To test whether there is a relationship between the
age of a member and his or her choice of sport, 643 members of the sports club are
randomly selected. Conduct a test for independence.

Sport 18 - 25 26 - 30 31 - 40 41 and over
racquetball 42 58 30 46
tennis 58 76 38 65
swimming 72 60 65 33

Solution:

c6

e25.21

f0.0003

hDecision: Reject null

Exercise:

Problem:

A major food manufacturer is concerned that the sales for its skinny French fries have
been decreasing. As a part of a feasibility study, the company conducts research into the
types of fries sold across the country to determine if the type of fries sold is independent
of the area of the country. The results of the study are below. Conduct a test for
independence.

Type of Fries Northeast South Central West
skinny fries 70 50 20 25
curly fries 100 60 15 30
steak fries 20 40 10 10
Exercise:
Problem:

According to Dan Lenard, an independent insurance agent in the Buffalo, N.Y. area, the
following is a breakdown of the amount of life insurance purchased by males in the
following age groups. He is interested in whether the age of the male and the amount of
life insurance purchased are independent events. Conduct a test for independence.

Age of < $200,000 - $401,001 - $1,000,000
Males None $200,000 $400,000 $1,000,000 +

20 - 29 40 15 40 0 fs)

Age of < $200,000 - $401,001 - $1,000,000

Males None $200,000 $400,000 $1,000,000 +

30 - 39 35 5 20 20 10

40 - 49 20 0 30 0 30

50 + 40 30 15 15 10
Solution:

e cl2

e e125.74

e f0

e hDecision: Reject null

Exercise:
Problem:
Suppose that 600 thirty—year—olds were surveyed to determine whether or not there is a

relationship between the level of education an individual has and salary. Conduct a test
for independence.

Not a high
Annual school High school College Masters or
Salary graduate graduate graduate doctorate
7 15 25 10 5
$30,000
$30,000 -
$40,000 20 40 70 30
$40,000 -
$50,000 10 20 40 55
eae 10 20 60

$60,000

Not a high

Annual school High school College Masters or
Salary graduate graduate graduate doctorate
ome | x0 5 10 150
Exercise:
Problem:

A Psychologist is interested in testing whether there is a difference in the distribution of
personality types for business majors and social science majors. The results of the study
are shown below. Conduct a Test of Homogeneity. Test at a 5% level of significance.

Open Conscientious Extrovert Agreeable Neurotic
Business 41 52 46 61 58
oe 72 75 63 80 65
Science
Solution:
e <4
e dChi-Square with df = 4
e e3.01
e fp-value = 0.5568
eh

ii. Do not reject the null hypothesis.
iv. There is insufficient evidence to conclude that the distribution of personality
types is different for business and social science majors.

Exercise:
Problem:
Do men and women select different breakfasts? The breakfast ordered by randomly

selected men and women at a popular breakfast place is shown below. Conduct a test of
homogeneity. Test at a 5% level of significance

French Toast Pancakes Waffles Omelettes
Men 47 35 28 53

Women 65 59 55 60

Solution:

e 63

e e4.01

e fp-value = 0.2601

eh
ii. Do not reject the null hypothesis.
iv. There is insufficient evidence to conclude that the distribution of breakfast
ordered is different for men and women.

Exercise:

Problem:

Is there a difference between the distribution of community college statistics students and
the distribution of university statistics students in what technology they use on their
homework? Of the randomly selected community college students 43 used a computer,
102 used a calculator with built in statistics functions, and 65 used a table from the
textbook. Of the randomly selected university students 28 used a computer, 33 used a
calculator with built in statistics functions, and 40 used a table from the textbook.
Conduct an appropriate hypothesis test using a 0.05 level of significance.

Solution:

° c2

e e7.05

¢ fp-value = 0.0294

eh
ii. Reject the null hypothesis.
iv. There is sufficient evidence to conclude that the distribution of technology use
for statistics homework is not the same for statistics students at community colleges
and at universities.

Exercise:

Problem:

A fisherman is interested in whether the distribution of fish caught in Green Valley Lake
is the same as the distribution of fish caught in Echo Lake. Of the 191 randomly selected
fish caught in Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 were
bass, and 24 were catfish. Of the 293 randomly selected fish caught in Echo Lake, 115
were rainbow trout, 58 were other trout, 67 were bass, and 53 were catfish. Perform the
hypothesis test at a 5% level of significance.

Solution:

¢ 3

e dChi-Square with df = 3

e e11.75

¢ fp-value = 0.0083

eh
ii. Reject the null hypothesis.
iv. There is sufficient evidence to conclude that the distribution of fish in Green
Valley Lake is not the same as the distribution of fish in Echo Lake.

Exercise:

Problem:

A plant manager is concerned her equipment may need recalibrating. It seems that the
actual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard
deviation should be at most + oz. In order to determine if the machine needs to be
recalibrated, 84 randomly selected boxes of cereal from the next day’s production were
weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be
recalibrated?

Solution:

e 83

¢ dChi-Square with df = 83

e e96.81

¢ fp-value = 0.1426; There is a 0.1426 probability that the sample standard deviation
is 0.54 or more.

¢ hDecision: Do not reject null; Conclusion: There is insufficient evidence to
conclude that the standard deviation is more than 0.5 oz. It cannot be determined
whether the equipment needs to be recalibrated or not.

Exercise:

Problem:

Consumers may be interested in whether the cost of a particular calculator varies from
store to store. Based on surveying 43 stores, which yielded a sample mean of $84 and a
sample standard deviation of $12, test the claim that the standard deviation is greater than
S15;

Exercise:

Problem:

Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for
her time to run the 7 % mile race is at most 3 minutes. To test her claim, Rupinder looks
up 5 of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57
minutes.

Solution:

e c4

e dChi-Square with df = 4

e e4.52

¢ £0.3402

e hDecision: Do not reject null.

Exercise:

Problem:

Airline companies are interested in the consistency of the number of babies on each
flight, so that they have adequate safety equipment. They are also interested in the
variation of the number of babies. Suppose that an airline executive believes the average
number of babies on flights is 6 with a variance of 9 at most. The airline conducts a
survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample
standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief.

Exercise:

Problem:

The number of births per woman in China is 1.6 down from 5.91 in 1966 (SourceWorld
Bank, 6/5/12). This fertility rate has been attributed to the law passed in 1979 restricting
births to one per woman. Suppose that a group of students studied whether or not the
standard deviation of births per woman was greater than 0.75. They asked 50 women
across China the number of births they had. Below are the results. Does the students’
survey indicate that the standard deviation is greater than 0.75?

# of births Frequency

0 5

iL 30

2 10

3 5
Solution:

e c49

¢ dChi-Square with df = 49

e e54.37

¢ fp-value = 0.2774; If the null hypothesis is true, there is a 0.2774 probability that the
sample standard deviation is 0.79 or more.

e hDecision: Do not reject null; Conclusion: There is insufficient evidence to
conclude that the standard deviation is more than 0.75. It cannot be determined if the
standard deviation is greater than 0.75 or not.

Exercise:

Problem:

According to an avid aquariest, the average number of fish in a 20—gallon tank is 10, with
a standard deviation of 2. His friend, also an aquariest, does not believe that the standard
deviation is 2. She counts the number of fish in 15 other 20—gallon tanks. Based on the
results that follow, do you think that the standard deviation is different from 2? Data: 11;
10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11

Exercise:

Problem:

The manager of "Frenchies" is concerned that patrons are not consistently receiving the
same amount of French fries with each order. The chef claims that the standard deviation
for a 10-ounce order of fries is at most 1.5 oz., but the manager thinks that it may be
higher. He randomly weighs 49 orders of fries, which yields a mean of 11 oz. anda
standard deviation of 2 oz.

Solution:
y) 2
°* ao” < (1.5)

e c48
e dChi-Square with df = 48

e e85.33
¢ f0.0007
e hDecision: Reject null.

Try these true/false questions.

Exercise:

Problem:

As the degrees of freedom increase, the graph of the chi-square distribution looks more
and more symmetrical.

Solution:
True
Exercise:
Problem: The standard deviation of the chi-square distribution is twice the mean.
Solution:

False
Exercise:

Problem:
The mean and the median of the chi-square distribution are the same if df = 24.
Solution:

False
Exercise:

Problem:

In a Goodness-of-Fit test, the expected values are the values we would expect if the null
hypothesis were true.

Solution:

True

Exercise:

Problem:

In general, if the observed values and expected values of a Goodness-of-Fit test are not
close together, then the test statistic can get very large and on a graph will be way out in
the right tail.

Solution:

True
Exercise:

Problem:
The degrees of freedom for a Test for Independence are equal to the sample size minus 1.
Solution:

False
Exercise:

Problem:

Use a Goodness-of-Fit test to determine if high school principals believe that students are
absent equally during the week or not.

Solution:
True

Exercise:

Problem: The Test for Independence uses tables of observed and expected data values.
Solution:

True
Exercise:

Problem:

The test to use when determining if the college or university a student chooses to attend
is related to his/her socioeconomic status is a Test for Independence.

Solution:

True

Exercise:

Problem: The test to use to determine if a six-sided die is fair is a Goodness-of-Fit test.
Solution:

True
Exercise:

Problem:

In a Test of Independence, the expected number is equal to the row total multiplied by the
column total divided by the total surveyed.

Solution:

True
Exercise:

Problem:

In a Goodness-of Fit test, if the p-value is 0.0113, in general, do not reject the null
hypothesis.

Solution:

False
Exercise:

Problem:

For a Chi-Square distribution with degrees of freedom of 17, the probability that a value
is greater than 20 is 0.7258.

Solution:

False
Exercise:

Problem:
If df = 2, the chi-square distribution has a shape that reminds us of the exponential.
Solution:

True

Review

This module provides an review on Chi-Square Distribution as a part of
Collaborative Statistics collection (col10522) by Barbara Illowsky and
Susan Dean.

The next two questions refer to the following real study:

A recent survey of U.S. teenage pregnancy was answered by 720 girls, age
12 - 19. 6% of the girls surveyed said they have been pregnant. (Parade
Magazine) We are interested in the true proportion of U.S. girls, age 12 -
19, who have been pregnant.

Exercise:

Problem:

Find the 95% confidence interval for the true proportion of U.S. girls,
age 12 - 19, who have been pregnant.

Solution:

(0.0424,0.0770)

Exercise:
Problem:
The report also stated that the results of the survey are accurate to
within + 3.7% at the 95% confidence level. Suppose that a new study
is to be done. It is desired to be accurate to within 2% of the 95%

confidence level. What is the minimum number that should be
surveyed?

Solution:

2401
Exercise:

Problem:

Given: X ~ Exp( Ji Sketch the graph that depicts: P(x > 1).

i
3

The next four questions refer to the following information:

Suppose that the time that owners keep their cars (purchased new) is
normally distributed with a mean of 7 years and a standard deviation of 2
years. We are interested in how long an individual keeps his car (purchased
new). Our population is people who buy their cars new.

Exercise:

Problem:
60% of individuals keep their cars at most how many years?
Solution:

75
Exercise:

Problem:

Suppose that we randomly survey one person. Find the probability that
person keeps his/her car less than 2.5 years.

Solution:

0.0122
Exercise:

Problem:

If we are to pick individuals 10 at a time, find the distribution for the
mean car length ownership.

Solution:

N(7,0.63)

Exercise:

Problem:

If we are to pick 10 individuals, find the probability that the sum of
their ownership time is more than 55 years.

Solution:

0:9911

Exercise:

Problem: For which distribution is the median not equal to the mean?

e AUniform

e BExponential
e CNormal

e DStudent-t

Solution:

B
Exercise:

Problem:

Compare the standard normal distribution to the student-t distribution,
centered at 0. Explain which of the following are true and which are
false.

e aAs the number surveyed increases, the area to the left of -1 for
the student-t distribution approaches the area for the standard
normal distribution.

e bAs the degrees of freedom decrease, the graph of the student-t
distribution looks more like the graph of the standard normal
distribution.

e clf the number surveyed is 15, the normal distribution should
never be used.

Solution:

e alrue
e bFalse
e cFalse

The next five questions refer to the following information:

We are interested in the checking account balance of a twenty-year-old
college student. We randomly survey 16 twenty-year-old college students.
We obtain a sample mean of $640 and a sample standard deviation of $150.
Let X = checking account balance of an individual twenty year old college
student.

Exercise:

Problem: Explain why we cannot determine the distribution of X.
Exercise:

Problem:

If you were to create a confidence interval or perform a hypothesis test

for the population mean checking account balance of 20-year old
college students, what distribution would you use?

Solution:

student-t with df = 15
Exercise:

Problem:

Find the 95% confidence interval for the true mean checking account
balance of a twenty-year-old college student.

Solution:

(560.07,719.93)
Exercise:
Problem:

What type of data is the balance of the checking account considered to
be?

Solution:

quantitative - continuous
Exercise:

Problem:

What type of data is the number of 20 year olds considered to be?

Solution:

quantitative - discrete
Exercise:
Problem:
On average, a busy emergency room gets a patient with a shotgun

wound about once per week. We are interested in the number of
patients with a shotgun wound the emergency room gets per 28 days.

e aDefine the random variable X.

e bState the distribution for X.

e cFind the probability that the emergency room gets no patients
with shotgun wounds in the next 28 days.

Solution:

¢ b P(4)
© 0.0183

The next two questions refer to the following information:

The probability that a certain slot machine will pay back money when a
quarter is inserted is 0.30 . Assume that each play of the slot machine is
independent from each other. A person puts in 15 quarters for 15 plays.
Exercise:

Problem:

Is the expected number of plays of the slot machine that will pay back
money greater than, less than or the same as the median? Explain your
answer.

Solution:

greater than
Exercise:

Problem:

Is it likely that exactly 8 of the 15 plays would pay back money?
Justify your answer numerically.

Solution:

No; P(x = 8) = 0.0348

Exercise:

Problem: A game is played with the following rules:

¢ it costs $10 to enter

e a fair coin is tossed 4 times

¢ if you do not get 4 heads or 4 tails, you lose your $10

e if you get 4 heads or 4 tails, you get back your $10, plus $30 more

Over the long run of playing this game, what are your expected
earnings?

Solution:

You will lose $5
Exercise:
Problem:
e The mean grade on a math exam in Rachel’s class was 74, with a
standard deviation of 5. Rachel earned an 80.
e The mean grade on a math exam in Becca’s class was 47, witha
standard deviation of 2. Becca earned a 51.

e The mean grade on a math exam in Matt’s class was 70, with a
standard deviation of 8. Matt earned an 83.

Find whose score was the best, compared to his or her own class.
Justify your answer numerically.
Solution:

Becca

The next two questions refer to the following information:

A random sample of 70 compulsive gamblers were asked the number of
days they go to casinos per week. The results are given in the following
graph:

Relative Frequency

0.3
0.2
0.1
1 2 3 4 5 6 7
Number of Days
Exercise:

Problem: Find the number of responses that were “5".

Solution:

14
Exercise:
Problem:

Find the mean, standard deviation, the median, the first quartile, the
third quartile and the IQR.

Solution:

e Sample mean = 3.2

e Sample standard deviation = 1.85
e Median = 3

e Quartile 1 = 2

¢ Quartile 3=5
e IQR=3

Exercise:

Problem:

Based upon research at De Anza College, it is believed that about 19%
of the student population speaks a language other than English at
home.

Suppose that a study was done this year to see if that percent has
decreased. Ninety-eight students were randomly surveyed with the
following results. Fourteen said that they speak a language other than
English at home.

e aState an appropriate null hypothesis.

e bState an appropriate alternate hypothesis.

e cDefine the Random Variable, P’.

e dCalculate the test statistic.

e eCalculate the p-value.

e fAt the 5% level of decision, what is your decision about the null
hypothesis?

e gWhat is the Type I error?

e¢ hWhat is the Type II error?

Solution:

e dz=-1.19
e e0.1171
e fDo not reject the null

Exercise:

Problem:

Assume that you are an emergency paramedic called in to rescue
victims of an accident. You need to help a patient who is bleeding
profusely. The patient is also considered to be a high risk for
contracting AIDS. Assume that the null hypothesis is that the patient
does not have the HIV virus. What is a Type I error?

Solution:

We conclude that the patient does have the HIV virus when, in fact, the
patient does not.

Exercise:

Problem:

It is often said that Californians are more casual than the rest of
Americans. Suppose that a survey was done to see if the proportion of
Californian professionals that wear jeans to work is greater than the
proportion of non-Californian professionals. Fifty of each was
surveyed with the following results. 15 Californians wear jeans to
work and 6 non-Californians wear jeans to work.

e C = Californian professional
e NC = non-Californian professional

e aState appropriate null and alternate hypotheses.

e bDefine the Random Variable.

e cCalculate the test statistic and p-value.

e dAt the 5% significance level, what is your decision?
e eWhat is the Type I error?

e fWhat is the Type II error?

Solution:

C27 > 221+) => 0.0136
e dReject the null

e eWe conclude that the proportion of Californian professionals that
wear jeans to work is greater than the proportion of non-
Californian professionals when, in fact, it is not greater.

e fWe cannot conclude that the proportion of Californian
professionals that wear jeans to work is greater than the
proportion of non-Californian professionals when, in fact, it is
greater.

The next two questions refer to the following information:

A group of Statistics students have developed a technique that they feel will
lower their anxiety level on statistics exams. They measured their anxiety
level at the start of the quarter and again at the end of the quarter. Recorded
is the paired data in that order: (1000, 900); (1200, 1050); (600, 700);
(1300, 1100); (1000, 900); (900, 900).

Exercise:

Problem: This is a test of (pick the best answer):

e Alarge samples, independent means
e¢ Bsmall samples, independent means
e Cdependent means

Solution:

G

Exercise:

Problem: State the distribution to use for the test.

Solution:

ts

Solution Sheet: Hypothesis Testing for Single Mean and Single Proportion
This module provides a solution sheet for the Hypothesis Testing: Single
Mean and Single Proportion chapter of the Collaborative Statistics
textbook/collection.

Class Time:
Name:

e aH:

e bH.:

e clIn words, CLEARLY state what your random variable X or P
represents.

e dState the distribution to use for the test.

e eWhat is the test statistic?

e fWhat is the p-value? In 1 — 2 complete sentences, explain what the p-
value means for this problem.

e gUse the previous information to sketch a picture of this situation.
CLEARLY, label and scale the horizontal axis and shade the region(s)
corresponding to the p-value.

e hIndicate the correct decision (“reject” or “do not reject” the null
hypothesis), the reason for it, and write an appropriate conclusion,
using complete sentences.

iAlpha:

iiDecision:

iiiReason for decision:
ivConclusion:

o Oo 0 90

e iConstruct a 95% Confidence Interval for the true mean or proportion.
Include a sketch of the graph of the situation. Label the point estimate
and the lower and upper bounds of the Confidence Interval.

Solution Sheet: Hypothesis Testing for Two Means, Paired Data, and Two
Proportions

This module provides a solution sheet for the Hypothesis Testing: Two
Means, Paired Data, Two Proportions chapter of the Collaborative Statistics
textbook/collection.

Class Time:
Name:

e aH:

e bH.:

e cin words, clearly state what your random variable X X ,
| os P -or Xq represents.

e dState the distribution to use for the test.

e eWhat is the test statistic?

e fWhat is the p-value? In 1 — 2 complete sentences, explain what the p-
value means for this problem.

e gUse the previous information to sketch a picture of this situation.
CLEARLY label and scale the horizontal axis and shade the region(s)
corresponding to the p-value.

e hIndicate the correct decision (“reject” or “do not reject” the null
hypothesis), the reason for it, and write an appropriate conclusion,
using complete sentences.

iAlpha:

iiDecision:

iiiReason for decision:
ivConclusion:

o Oo 0 0

e iln complete sentences, explain how you determined which
distribution to use.

Solution Sheet: The Chi-Square Distribution
This module provides a solution sheet for the Chi-Square Distribution
chapter of the Collaborative Statistics textbook/collection.

Class Time:
Name:

e aH:

e bH.:

e cWhat are the degrees of freedom?

e dState the distribution to use for the test.

e eWhat is the test statistic?

e fWhat is the p-value? In 1 — 2 complete sentences, explain what the p-
value means for this problem.

e gUse the previous information to sketch a picture of this situation.
Clearly label and scale the horizontal axis and shade the region(s)
corresponding to the p-value.

e hIndicate the correct decision (“reject” or “do not reject” the null
hypothesis) and write appropriate conclusions, using complete
sentences.

iAlpha:

iiDecision:

iiiReason for decision:
ivConclusion:

o Oo 0 90

Symbols and their Meanings
This module defines symbols used throughout the Collaborative Statistics
textbook.

Chapter
(1st used) Symbol Spoken Meaning
Sampling The square root
and Data Vv of os
Sampling Pj pee :
Sacha 1 i (a specific
number)
Descriptive the first
Statistics Ql uae oue quartile
Descriptive the second
Statistics Q2 Qamewe quartile
Descriptive : the third
Statistics Q3 ae tae quartile
Descriptive IQR inter-quartile Q3-
Statistics range Q1i=IQR
Descriptive “ “ba sample
Statistics mean
Descriptive P ani population
Statistics mean

Chapter
(1st used)

Descriptive
Statistics

Descriptive
Statistics

Descriptive
Statistics

Descriptive
Statistics

Descriptive
Statistics

Probability
Topics

Probability
Topics

Probability
Topics

Probability
Topics

Symbol

S Sy SX

O Ox OX

{}

Spoken

s-squared

sigma

sigma-squared

capital sigma

brackets

Event A

probability of
A

Meaning
sample
standard

deviation

sample
variance

population
standard

deviation

population
variance

sum

set notation

sample
space

event A

probability
of A
occurring

Chapter
(1st used)

Probability
Topics

Probability
Topics

Probability
Topics

Probability
Topics

Probability
Topics

Probability
Topics

Probability
Topics

Discrete
Random
Variables

Symbol

P(A| B)

P (Aor B)

P (Aand B)

A?

PDF

Spoken

probability of
A given B

prob. of A or B

prob. of A and
B

A-prime,
complement of
A

prob. of
complement of
A

green on first
pick

prob. of green
on first pick

prob.
distribution
function

Meaning

prob. of A
occurring
given B has
occurred

prob. of A
or B or both
occurring
prob. of
both A and

B occurring
(same time)

complement
of A, notA

same

same

same

same

Chapter
(1st used)

Discrete
Random
Variables

Discrete
Random
Variables

Discrete
Random
Variables

Discrete
Random
Variables

Discrete
Random
Variables

Discrete
Random
Variables

Discrete
Random
Variables

Discrete
Random
Variables

Symbol

IV

Spoken

the distribution

of X

binomial

distribution

geometric

distribution

hypergeometric

dist.

Poisson dist.

Lambda

greater than or

equal to

Meaning

the random
variable X

same

same

same

same

same

average of

Poisson
distribution

same

Chapter
(1st used)

Discrete
Random
Variables

Discrete
Random
Variables

Discrete
Random
Variables

Continuous
Random
Variables

Continuous
Random
Variables

Continuous
Random
Variables

Continuous
Random
Variables

Continuous
Random
Variables

Symbol

IA

Exp

Spoken

less than or
equal to

equal to

not equal to

f of x

prob. density
function

uniform
distribution

exponential
distribution

Meaning

same

same

same

function of
x

same

same

same

critical
value

Chapter
(1st used)

Continuous
Random
Variables

Continuous
Random
Variables

The Normal
Distribution

The Normal
Distribution

The Normal
Distribution

The Central
Limit
Theorem

The Central
Limit
Theorem

The Central
Limit
Theorem
The Central
Limit
Theorem

CLT

Le

Me

Spoken

f of x equals

normal
distribution

Z-SCOre

standard
normal dist.

Central Limit

Theorem

X-bar

mean of X

mean of X-bar

Meaning

same

decay rate
(for exp.
dist.)

same

same

same

same

the random
variable X-
bar

the average
of X

the average
of X-bar

Chapter
(1st used)

The Central
Limit
Theorem

The Central
Limit
Theorem

The Central
Limit
Theorem
The Central
Limit
Theorem

Confidence
Intervals

Confidence
Intervals

Confidence
Intervals

Confidence
Intervals

Confidence
Intervals

Confidence
Intervals

Symbol

Or

Or

dX

CL

CI

EKBM

df

Spoken

standard
deviation of X

standard
deviation of X-
bar

sum of X

sum of x
confidence
level

confidence
interval

error bound for
a mean

error bound for
a proportion

student-t
distribution

degrees of
freedom

Meaning

same

same

same

same

same

same

same

same

same

same

Chapter
(1st used)

Confidence
Intervals

Confidence
Intervals

Confidence
Intervals

Hypothesis
Testing

Hypothesis
Testing

Hypothesis
Testing

Hypothesis
Testing

Hypothesis
Testing

Hypothesis
Testing

Ay,

X1—X2

Spoken

student-t with
a/2 area in
right tail

p-prime; p-hat

q-prime; q-hat

H-naught, H-
sub 0

H-a, H-sub a

H-1, H-sub 1

alpha

beta

X1-bar minus
X2-bar

Meaning

same

sample
proportion
of success

sample
proportion
of failure

null
hypothesis

alternate
hypothesis

alternate
hypothesis

probability
of Type I
error

probability
of Type II
error

difference
in sample
means

Chapter
(1st used)

Chi-Square
Distribution

Linear
Regression
and
Correlation

Symbol

1 — P2
P’,—P”
P1 — P2
x2

O

E
y=a+bx
y

r

Spoken

mu-1 minus
mu-2

P1-prime
minus P2-
prime

pl minus p2

Ky-square

Observed

Expected

y equals a plus
b-x

y-hat

correlation
coefficient

Meaning

difference
in
population
means

difference
in sample
proportions

difference
in

population
proportions

Chi-square

Observed
frequency

Expected
frequency

equation of
a line

estimated
value of y

same

Chapter

(1st used) Symbol
E
SSE
1.9s

F-

Distribution F

and

ANOVA

Symbols and their Meanings

Spoken
error

Sum of
Squared Errors

1.9 times s

F-ratio

Meaning

same

same

cut-off
value for
outliers

F ratio

Formulas

This module provides an overview of Statistics Formulas used as a part of
Collaborative Statistics collection (col10522) by Barbara Illowsky and
Susan Dean.

Formula

Factorial

n! = n(n — 1)(n — 2)... (1)

O!=1
Formula
Combinations

=e

Formula
Binomial Distribution

X~B(n,p)

PI XB! pg © Mone 20,4 2 on05
Formula
Geometric Distribution

X~-G(p)

21 0, Ge x) = 9° p fore = 1.9. 3.33
Formula
Hypergeometric Distribution

X~H (r,b,n)
a)
y cdi WO Gens se (te ;
(=*) = ("5
Formula

Poisson Distribution

X~P (u)

P(X=2)=4>

z!
Formula
Uniform Distribution

X~U(a, b)

f(X) =po,a<2<b
Formula

Exponential Distribution
X ~ Exp(m)

{= me" ame Oe 0
Formula
Normal Distribution

X~N(p, 07)

1 cea?
fix) =——e ww , -w<2< 0
oV or

Formula
Gamma Function

DZ \af, ee dee ee

r(4) = va

I'(m +1) = m! for m, a nonnegative integer
otherwise: (a + 1) = aI'(a)

Formula

Student-t Distribution

X ~tae¢

Z-N(0,1) , Y-X2, ,n = degrees of freedom
Formula
Chi-Square Distribution

X~X?2,

n-2 —-2
Vi () = Era) ,x > 0, n= positive integer and degrees of freedom
2

Formula
F Distribution

X ~Fat(n),df(d)
df(n) =degrees of freedom for the numerator

df(d) =degrees of freedom for the denominator

Y,
Pe — We > Y, W are chi-square

Notes for the TI-83, 83+, 84 Calculator
Notes and tips for using TI-83, TI-83+, and TI-84 calculators for statistics
applications.

Quick Tips
Legend
. fi }

represents a button press
e [| | represents yellow command or green letter behind a key
e < >represents items on the screen

To adjust the contrast
Press

| 2nd |
, then hold

to increase the contrast or

to decrease the contrast.

To capitalize letters and words
Press

ALPHA)

to get one capital letter, or press

, then

ALPHA]

to set all button presses to capital letters. You can return to the top-level
button values by pressing

(ALPHA)
again.

To correct a mistake
If you hit a wrong button, just hit

and start again.

To write in scientific notation
Numbers in scientific notation are expressed on the TI-83, 83+, and 84
using E notation, such that...

© 4321E4=4.321 x 104
e 4.321 E-4= 4.321 x 10-4

To transfer programs or equations from one calculator to another:
Both calculators: Insert your respective end of the link cable cable and
press

, then [LINK].
Calculator receiving information:

Use the arrows to navigate to and select<RECEIVE>
Press

Calculator sending information:

Press appropriate number or letter.

Use up and down arrows to access the appropriate item.
Press to select item to transfer.

Press right arrow to navigate to and select<TRANSMIT>.
Press

Note:ERROR 35 LINK generally means that the cables have not been
inserted far enough.

Both calculators: Insert your respective end of the link cable cable Both
calculators: press

| 2nd
, then [QUIT ] To exit when done.

Manipulating One-Variable Statistics

Note: These directions are for entering data with the built-in statistical
program.

Data Frequency

-2 10

Data Frequency

-1 3
0 4
1 5
3 8

Sample DataWe are manipulating 1-variable statistics.
To begin:
Turn on the calculator.

Access Statistics mode.

STAT

Select<4:C1lrList>to clear data from lists, if desired.

Enter list [ L1]to be cleared.

Display last instruction.

, [ENTRY]
Continue clearing remaining lists in the same fashion, if desired.

GB
| 2nd |
, eae,
Access statistics mode.
STAT
Select@ i Gale =
ENTER]
Enter [L1]. (You may [L1])
data. Data need to o Type in a data value and enter it.
values go arrow over (For negative numbers, use the
into to negate (-) key at the bottom of the

keypad)

=)

ans)

’

ENTER)

o Continue in the same manner until
all data values are entered.

In[ L2 ], enter the (eat).
frequencies for o Type in a frequency and enter it. (If a
each data value in data value appears only once, the

frequency is "1")

eae)

ENTER]

o Continue in the same manner until all
data values are entered.

Access Statistics mode.

STAT
Navigate to<CALC>
Access<1:1-var Stats>

ENTER)

Indicate that the data is in[ L1 ]...

Exp
,» [LI],

..and indicate that the frequencies are in[ L2 }.

| 2nd |
» [L2],

The statistics should be displayed. You may arrow down to get remaining
Statistics. Repeat as necessary.

Drawing Histograms

Note:We will assume that the data is already entered

We will construct 2 histograms with the built-in STATPLOT application.
The first way will use the default ZOOM. The second way will involve
customizing a new graph.

Access graphing mode.

, [STAT PLOT]
Select<1: plot 1>To access plotting - first graph.

Gi

Use the arrows navigate go to<ON>to turn on Plot 1.

<ON>.

ENTER]

Use the arrows to go to the histogram picture and select the histogram.
ENTER]

Use the arrows to navigate to<Xlist>
If "L1" is not selected, select it.

| 2nd |
,» [LI],

Gi

Use the arrows to navigate to<Freq>.
Assign the frequencies to[ L2 ].

| 2nd |
» [L2],

ENTER]

Go back to access other graphs.

, (stat PEO]
Use the arrows to turn off the remaining plots.
Be sure to deselect or clear all equations before graphing.

To deselect equations:

Access the list of equations.

Select each equal sign (=).

GA
Cs
ENTER]

Continue, until all equations are deselected.
To clear equations:

Access the list of equations.

Ys

Use the arrow keys to navigate to the right of each equal sign (=) and
clear them.

Repeat until all equations are deleted.
To draw default histogram:

Access the ZOOM menu.

CD

Select<9 : ZoomStat>

1)

The histogram will show with a window automatically set.

To draw custom histogram:

Access to set the graph parameters.
ie) Ain = —2.5
Oo A may = 3.0
°o Xs) = 1 (width of bars)
° Ynin = 0
° Ynax = 10
o Y;.) = 1 (spacing of tick marks on y-axis)
2 X res =1

Access to see the histogram.

To draw box plots:

Access graphing mode.

, [STAT PLOT]
Select<1:Plot 1>to access the first graph.

Gi

Use the arrows to select<ON>and turn on Plot 1.

ENTER]

Use the arrows to select the box plot picture and enable it.

Use the arrows to navigate to<Xlist>
If "L1" is not selected, select it.

2nd
a.

Use the arrows to navigate to<Freq>.
Indicate that the frequencies are in[ L2].

» [L2],

Go back to access other graphs.

, [STAT PLOT |
Be sure to deselect or clear all equations using the method
before graphing mentioned above.
View the box plot.

(GRAPH
, [STAT PLOT]

Linear Regression

Sample Data

The following data is real. The percent of declared ethnic minority students
at De Anza College for selected years from 1970 - 1995 was:

Year Student Ethnic Minority Percentage
1970 14.13
1973 12.27
1976 14.08
1979 18.16
1982 27.64
1983 28.72
1986 31.86
1989 33.14
1992 45.37
1995 93.1

The independent variable is "Year," while the independent variable is
"Student Ethnic Minority Percent."

Student Ethnic Minority Percentage

60 Stud. Ethnic Min. Perc.

Percent

1960 1970 1980 1990 2000

Year

By hand, verify the scatterplot above.

Note:The TI-83 has a built-in linear regression feature, which allows the
data to be edited. The x-values will be in

[L1]
; the y-values in

[L2]

To enter data and do linear regression:

ON Turns calculator on

Before accessing this program, be sure to turn off

all plots. o Access graphing

mada

11U0Ue.

, [STAT PLOT]
o Turn off all plots.

as)
Round to 3 decimal
places. To do so: © Access the mode menu.
| MODE |

, [STAT PLOT]
o Navigate to <Float> and then to the right
to <3>.

o All numbers will be rounded to 3 decimal
places until changed.

Enter statistics mode and clear lists[ L1 ]and[L2], as describe above.

Enter editing mode to insert values for x and y.

STAT
ENTER]
Enter each value. Press to continue.
Gi
To display the correlation coefficient:
Access the catalog.
| 2nd |
, [CATALOG ]
Arrow down and select<DlagnosticOn>
GS
ENTER]

ENTER]

STAT

Tandr2will be displayed during regression calculations.
Access linear regression.

STAT

Select the form ofy = a + bx

aud)
ENTER]
The display will show:
LinReg
°y=a-+ bz
e a = —3176.909
¢« b=1.617
er? = 0.924
« r= 0.961

This means the Line of Best Fit (Least Squares Line) is:

e y = —3176.909 + 1.6172
e Percent = —3176.909 + 1.617(year #)

The correlation coefficient r = 0.961
To see the scatter plot:

Access graphing mode.

| 2nd
, [STAT PLOT]

Select<1: plot 1>To access plotting - first graph.

Navigate and select<ON>to turn on Plot 1.
<ON>

Navigate to the first picture.
Select the scatter plot.

ENTER]

Navigate to<Xlist>
If[L1]is not selected, press ,| L1 ]to select it.
L 2nd |

Confirm that the data values are in[ L1].
<ON>

ENTER]

Navigate to<Ylist>
Select that the frequencies are in[ L2 ].

» [L2],

Go back to access other graphs.

ro rar ni aAwTT

, SiReUA
Use the arrows to turn off the remaining plots.
Access___to set the graph

parameters. ° Xmin = 1970

Oo XK way = 2000

© Xs = 10 (spacing of tick marks on
x-axis)

& Yan = 0.05

2 Vane = 60

o Y,. = 10 (spacing of tick marks on
y-axis)

2 X yes =1

Be sure to deselect or clear all equations before graphing, using the
instructions above.
Press to see the scatter plot.

To see the regression graph:

Access the equation menu. The regression equation will be put into
YI:

Y=

Access the vars menu and navigate to<5: Statistics>

Navigate to<EQ>.
E<allee contains the regression equation which will be entered in
RegEQ> bak

Gi

Press@Z). The regression line will be superimposed over scatter plot.

To see the residuals and use them to calculate the critical point for an
outlier:

Access the list. RESID will be an item on the menu.
Navigate to it.

, erSail,

<RESID>
Confirm twice to view the list of residuals. Use the arrows to select
them.

ENTER]
Gi
The critical point for an 1.9V 2= where:
outlier is: o m=number of pairs of
data
o SSE = sum of the squared
errors

o S~ residual?

Store the residuals in[ L3 ].

STOP

Calculate the (residual)? _ Note that. — 2 = 8
n—2

Store this value in| L4 ].

STOP
| 2nd
» ey

Calculate the critical value using the equation above.

E E

wy wy wy
-

2:

Verify that the calculator displays: 7.642669563. This is the critical value.

Compare the [L3]to 7.64. If the absolute value is greater than 7.64,

absolute value of then the (x, y) corresponding point is an outlier. In
each residual this case, none of the points is an outlier.
value in

To obtain estimates of y for various x-values:
There are various ways to determine estimates for "y". One way is to
substitute values for "x" in the equation. Another way is to use the

on the graph of the regression line.
TI-83, 83+, 84 instructions for distributions and tests

Distributions
Access DISTR (for Distributions").

For technical assistance, visit the Texas Instruments website at
http://www.ti.com and enter your calculator model into the "search" box.
Binomial Distribution

e binompdf(n,p, X) corresponds to P(X = x)

e binomcdf(n,p, X) corresponds to P(X < x)

e To see a list of all probabilities for x: 0,1,...,n, leave off the "x"
parameter.

Poisson Distribution

e poissonpdf(A, xX) corresponds to P(X = x)
e poissoncdf (A, xX) corresponds to P(X < x)

Continuous Distributions (general)

e —oo uses the value -1EE99 for left bound
¢ oo uses the value 1EE99 for right bound

Normal Distribution

normalpdf(x,U,0) yields a probability density function value
(only useful to plot the normal curve, in which case "xX" is the variable)
normalcdf(left bound, right bound, U,©o) corresponds
to P(left bound < X < right bound)

normalcdf(left bound, right bound) corresponds to
P(left bound < Z < right bound) - standard normal

invNorm(p,U, 0) yields the critical value, k: P(X < k) =p
invNorm(p) yields the critical value, k: P(Z < k) = p for the
standard normal

Student-t Distribution

e tpdf(x, df) yields the probability density function value (only

useful to plot the student-t curve, in which case "X" is the variable)

e tcdf(left bound, right bound, df) corresponds to P(left

bound < t < right bound)

Chi-square Distribution

e X*pdf(x, df) yields the probability density function value (only

useful to plot the chi curve, in which case "Xx" is the variable)

¢ X*cdf(left bound, right bound, df) corresponds to

P(left bound < X? < right bound)

F Distribution

e Fpdf(x,dfnum, dfdenom) yields the probability density function

value (only useful to plot the F curve, in which case "X" is the
variable)

e Fcdf(left bound, right bound, dfnum, dfdenom)

corresponds to P(left bound < F < right bound)

Tests and Confidence Intervals

Access STAT and TESTS.

For the Confidence Intervals and Hypothesis Tests, you may enter the data
into the appropriate lists and press DATA to have the calculator find the
sample means and standard deviations. Or, you may enter the sample means
and sample standard deviations directly by pressing STAT once in the
appropriate tests.

Confidence Intervals

e ZInterval is the confidence interval for mean when o is known

e TInterval is the confidence interval for mean when o is unknown;
s estimates o.

e 1-PropZInt is the confidence interval for proportion

Note:The confidence levels should be given as percents (ex. enter "95" or
"95" for a 95% confidence level).

Hypothesis Tests

e Z-TeSt is the hypothesis test for single mean when o is known

e T-Test is the hypothesis test for single mean when o is unknown; s
estimates o.

e 2-SampZTest is the hypothesis test for 2 independent means when
both o's are known

e 2-SampTTest is the hypothesis test for 2 independent means when
both o's are unknown

e 1-PropZTest is the hypothesis test for single proportion.

e 2-PropZTest is the hypothesis test for 2 proportions.

e X*-Test is the hypothesis test for independence.

e X*GOF-Test is the hypothesis test for goodness-of-fit (TI-84+ only).

e LinRegTTEST is the hypothesis test for Linear Regression (TI-84+
only).

Note:Input the null hypothesis value in the row below "Inpt." For a test
of a single mean, "[1@" represents the null hypothesis. For a test of a single
proportion, "©" represents the null hypothesis. Enter the alternate
hypothesis on the bottom row.

Tables

Note:When you are finished with the table link, use the back button on
your browser to return here.

Tables (NIST/SEMATECH e-Handbook of Statistical Methods,
http://www. itl nist.gov/div898/handbook/, January 3, 2009)

Student-t table

Normal table

Chi-Square table

F-table

All four tables can be accessed by going to

http://www.itl nist. gov/div898/handbook/eda/section3/eda367.htm

95% Critical Values of the Sample Correlation Coefficient Table

95% Critical Values of the Sample Correlation Coefficient

Note:The url for this table is http://cnx.org/content/m17098/latest/